 # Wave Motion and Waves on a String

## Concept Of Physics

### H C Verma

1   A wave pulse passing on a string with a speed of $40 cm/s$ in the negative x-direction has its maximum at $x = 0$ at $t = 0$. Where will this maximum be located at $t = 5 s$ ?

##### Solution :

v= 40 cm/sec $\\$ As velocity of a wave is constant location of maximum after 5 sec $\\$ = 40 $\times$ 5 =200 cm along negative x-axis.

2   The equation of a wave travelling on a string stretched along the X-axis is given by $y = A e^-(\frac{x}{a} +\frac{t}{T})^2$ (a) Write the dimensions of A, a and T. (b) Find the wave speed. (c) In which direction is the wave. travelling ? (d) Where is the maximum of the pulse located at t T ? At t - 2 T ?

##### Solution :

Given $y = A e^-[(\frac{x}{a} +\frac{t}{T})]^2$ $\\$ a)$[A] = [M^0L^1T^0],[T] = [M^0L^0T^1]$ $\\$ $[A] = [M^0L^1T^0]$ $\\$ b) wave speed, $v = \frac {\lambda}{T} = \frac{a}{T}$ [wave length $\lambda = a$] $\\$ c) If $y =f(t- \frac{X}{V}) \Rightarrow$ Wave is travelling in positive direction and if $y = f(t + \frac{X}{V}) \Rightarrow$ Wave is travelling in negative direction $\\$ so, $y =A e^-[(\frac{x}{a} +\frac{t}{T})]^2 = Ae^-(\frac{1}{T}[\frac{X}{\frac{a}{T}} + t]^2$ $\\$ = $Ae^-(\frac{1}{T})[\frac{X}{V} + t]^2$ $\\$ i.e. $y$ = $f$ {$t + (\frac{X}{V})$} $\\$ D)Wave speed, $V= \frac{a}{T}$ $\\$ $\therefore$ Max. of pulse at $t = T$ is $(\frac{a}{T}) \times T =a$(negative X-axis) $\\$ Max. of plus at $t =2T$ is $(\frac{a}{T}) \times 2T =2a$(along negative X-axis) $\\$ So, the wave travels in negative X-direction

3   Figure $(15-E1)$ shows a wave pulse at $t = 0$. The pulse moves to the right with a speed of $10 cm/s$. Sketch the shape of the string at $t = 1 s, 2 s$ and $3 s$.

##### Solution :

At $t = 1$ sec, $\qquad$ $s_1 = vt =10 \times 1 = 10 cm$ $\\$ At $t = 2$ sec, $\qquad$ $s_1 = vt =10 \times 2 = 20 cm$ $\\$ At $t = 3$ sec, $\qquad$ $s_1 = vt =10 \times 3 = 30 cm$

4   A pulse travelling on a string is represented by the function $y = \frac{a^3}{(x - vt)^2+ a^2},$ where a = 5 mm and v = 20 cm/s. Sketch the shape of the string at t = 0, 1 s and 2 s. Take x = 0 in the middle of the string.

##### Solution :

The pulse is given by, $y$ = $[\frac{(a^3)}{ \big\{ (x - vt)^2+ a^2 \big\} }]$ $\\$ $a = 5mm = 0.5 cm, v= 20 cm/s$ $\\$ At t= 0s, y = $\frac{a^3}{(x^2 + a^2)}$ $\\$ The graph between y and x can be plotted by taking different values of x. $\\$ (left as exercise for the student) $\\$ Similarly, at t = 1 s, $y = [\frac{(a^3)}{ \big\{ (x - vt)^2+ a^2 \big\} }]$ $\\$ and at t=2 s,$\qquad$ $y = [\frac{(a^3)}{ \big\{ (x - vt)^2+ a^2 \big\} }]$

5   The displacement of the particle at x = 0 of a stretched string carrying a wave in the positive x-direction is given by $f(t) = A sin(t/7)$. The wave speed is v. Write the wave equation.

##### Solution :

6   The displacement of the particle at x = 0 of a stretched string carrying a wave in the positive x-direction is given by $f(t) = A sin(t/7)$. The wave speed is v. Write the wave equation.

##### Solution :

At X=0, $f(t) = a sin (\frac{t}{T})$ $\\$ Wave speed =v $\\$ $\Rightarrow \lambda = wavelength = vT (T=Time Period)$ $\\$ So, general equation of wave $\\$ $Y =A sin [(\frac{t}{T}) - (\frac{x}{vT})]$ $\qquad$ [because $y = f(\big(\frac{t}{T} - \frac{x}{\lambda}\big))$

7   A wave pulse is travelling on a string with a speed v towards the positive X-axis. The shape of the string at t = 0 is given by $g(x) = A sin(x/a)$, where A and a are constants. (a) What are the dimensions of A and a ? (b) Write the equation of the wave for a general time t, if the wave speed is v.

##### Solution :

At =0, $g(x) =A sin (\frac{x}{a})$ $\\$ a) $[M^0L^1T^0] =[L]$ $\\$ $\qquad a = [M^0L^1T^0] =[L]$$\\$ b)Wave speed = v $\\$ $\therefore$ Time period, $T= \frac{a}{v} (a= wave length = \lambda)$ $\\$ $\therefore$ General equation of wave $\\$ $y = A \ sin \big\{ (x/a)- t/(a/v) \big\}$ $\\$ = $A \ sin \big\{\frac {(x - vt)}{a}\big\}$

8   A wave propagates on a string in the positive x-direction at a velocity v. The shape of the string at t = to is given by g(x, to) = $A sin(x/a)$ . Write the wave equation for a general time t.

##### Solution :

At $t = t_0, g(x,t_0) = A \ sin(x/a)$ ...........(1) $\\$ For a wave traveling in the positive x-direction, the general equation is given by $y= f \big(\frac{x}{a} - \frac{t}{T} \big)$ $\\$ Putting $t = - t_0$ and comparing with equation (1), we get $\\$ $\Rightarrow g(x,0)= A \ sin \big\{ (x/a) + (t_0/T) \big\}$ $\\$ $\Rightarrow g(x,0)= A \ sin \big\{ (x/a) + (t_0/T) - (t/T) \big\}$ $\\$ As t = a/v (a= wave length, v= speed of the wave) $\\$ $\Rightarrow y = Asin \big(\frac{x}{a} + \frac{t_0}{a/v} - \frac{t}{a/v} \big)$ $\\$ =$Asin \big( \frac{x + v(t_0 - t)}{a} \big)$ $\\$ $\Rightarrow y = Asin \big[ \frac{x - v(t - t_0)}{a} \big]$

9   The equation of a wave travelling on a string is $y = (0.10 mm) sin[(31.4 m^{-1})x + (314 s^{-1})t]$. (a) In which direction does the wave travel ? (b) Find the wave speed, the wavelength and the frequency of the wave. (c) What is the maximum displacement and the maximum speed of a portion of the string ?

##### Solution :

The equation of the wave is given by $\\$ $y = (0.10 mm) sin[(31.4 m^{-1})x + (314 s^{-1})t] \qquad y= r \ sin { (2 \pi x / \lambda)} +wt)$ $\\$ a)Negative x-direction b)k = 31.4 $m^{-1}$ $\\$ $\Rightarrow 2 \lambda /\lambda =3.14 \Rightarrow 2 \pi/3.14 = 0.2 mt = 20cm$ $\\$ Again, $w=314 s^{-1}$ $\\$ $\Rightarrow 2 \pi f = 314 \Rightarrow f = 314/2 \pi =314 / (2 \times (3/14)) = 50sec^-{1}$ $\\$ $\therefore wave speed, v = \lambda f = 20 \times 50 =1000 cm/s$ $\\$ c)Max. displacement = 0.10mm Max.velocity = aw = $0.1 \times 10^{-1} \times 314 = 3.14 cm/sec$.

10   A wave travels along the positive x-direction with a speed of 20 m/s. The amplitude of the wave is 0.20 cm and the wavelength 2.0 cm. (a) Write a suitable wave equation which describes this wave. (b) What is the displacement and velocity of the particle at x = 2.0 cm at time t = 0 according to the wave equation written ? Can you get different values of this quantity if the wave equation is written in a different fashion ?

##### Solution :

Wave speed, v=20 m/s $\\$ A = 0.20 cm $\\$ $\lambda = 2 cm$ $\\$ a)Equation of wave along the x-axis $\\$ $y= A \ sin (kx-wt)$ $\\$ $\therefore k = 2 \ pi / \lambda = 2 \ pi /2 = \pi cm^{-1}$ $\\$ $T= \lambda/V =2/2000 = 1/1000 sec = 10^{-3} sec$ $\\$ $\Rightarrow w = 2 \ pi/T = 2 \pi \times 10^{-3}sec$ $\\$ So, the wave equation is, $\\$ $\therefore y = (0.2 cm)sin \big[ (\pi cm^{-1})X -(2 \pi \times 10^{3} sec^{-1})t \big]$ $\\$ b)At $x=2cm, and t= 0$, $\\$ $y = (0.2cm) \ sin(\pi/2) = 0$ $\\$ $\therefore v = r w cos \pi x = 0.2 \times 2000 \pi \times cos 2 \pi=400 \pi$ $\\$ $\qquad = 400 \times (3.14) =1256 \ cm/s$ $\\$ $\qquad = 400 \pi \ cm/s = 4 \pi \ m/s$

11   A wave is described by the equation $y = (1.0 \ mm) sin \ \pi \big(\frac{x}{2.0 \ cm} - \frac{t}{0.01 \ s} \big)$ (a) Find the time period and the wavelength . (b) Write the equation for the velocity of the particles. Find the speed of the particle at x = 1.0 cm at time t = 0.01 s. (c) What are the speeds of the particles at x = 3'0 cm, 5.0 cm and 7.0 cm at t 0.01 s ? (d) What are the speeds of the particles at x 1.0 cm at t = 0.011, 0.012, and 0.013 s?

##### Solution :

$y = (1.0 \ mm) sin \ \pi \big(\frac{x}{2.0 \ cm} - \frac{t}{0.01 \ s} \big)$ $\\$ a)$4T = 2 \times 0.01 = 0.02 \ sec = 20 \ ms$ $\\$ $\lambda = 2 \times 2 = 4 \ cm$ $\\$ b)$v = dy/dt = d/dt[sin 2 \pi (x/4 - t/0.02)] = -cos 2 \pi \big\{x/4) - (t/0.02) \big\} \times 1/(0.02)$ $\\$ $\Rightarrow v= -50 \ cos 2 \pi \big\{(x/4) -(t/0.02) \big\}$ $\\$ $at x=1 and t=0.01 sec$ ,$v= -50 \ cos 2*[(1/4) - (1/2)] = 0$ $\\$ c) i) $at x= 3 cm, t =0.01$ sec $\\$ $v = -50 cos 2 \pi (3/4 - 1/2) =0$ $\\$ ii) $at x=5 cm, t= 0.01 sec, v=0$ (putting the values) $\\$ iii)$at x= 7 cm, t=0.01 sec, v=0$ $\\$ $v = -50 cos \ 2 \pi \big\{(1/4) -(0.011/0.02) \big\} = -50 cos (3 \pi /5) = -9.7 cm/sec$ $\\$ (Similarly other two can be calculated )

12   A particle on a stretched string supporting a travelling wave, takes 5.0 ms to move from its mean position to the extreme position. The distance between two consecutive particles, which are at their mean positions, is 2.0 cm. Find the frequency, the wavelength and the wave speed.

##### Solution :

Time period, $T= 4 \times 5 ms = 20 \times 10^{-3} = 2 \times 10^{-2} s$ $\\$ $\qquad \lambda = 2 \times 2 cm =4 cm$ $\\$ $frequency, f = 1/T = 1/(2\times 10^{-2}) = 50 s^{-1} = 50 Hz$ $\\$ $Wave \ speed = \lambda f = 4 \times 50 m/s =2000 m/s = 2 m/s$

13   Figure $(15-E2)$ shows a plot of the transverse displacements of the particles of a string at t = 0 through which a travelling wave is passing in the positive x-direction. The wave speed is $20 cm/s$. Find (a) the amplitude, (b) the wavelength, (c) the wave number and (d) the frequency of the wave.

##### Solution :

Given that, $v=200 m/s$ $\\$ a) Amplitude, $A = 1 mm$ $\\$ b)Wave length, $\lambda = 4 cm$ $\\$ c)Wave number, $n= 2 \pi / \lambda =(2 \times 3.14)/4 =1.57 cm^{-1}$ (wave number = k) $\\$ d) frequency, $f= 1/T = (26/ \lambda)/20 = 20/4 = 5Hz$ $\\$ (where time period $T= \lambda/v)$

14   A wave travelling on a string, at a speed of $10 m/s$ causes each particle of the string to oscillate with a time period of $20 ms$. (a) What is the wavelength of the wave ? (b) If the displacement of a particle is $1.5 mm$ at a certain instant, what will be the displacement of a particle $10 cm$ away from it at the same instant ?

##### Solution :

Wave speed = v= 10 m/sec $\\$ Time period = $T=20 ms = 20 \times 10^-{3} = 2 \times 10^{-2} sec$ $\\$ a)wave length, $\lambda = vT = 10 \times 2 \times 10^{-2} = 0.2 m =20cm$ $\\$ b)wave length,$\lambda = 20cm$ $\\$ $\therefore$ phase $diff^{n} = (2 \pi / \lambda) x = ( 2 \pi /20) \times 10 = \pi rad$ $\\$ $y_1 = a \ sin (wt - kx) \Rightarrow 1.5 = a \ sin(wt - kx)$ $\\$ So, the displacement of the particle at a distance x = 10cm. $\\$ $[\phi = \frac{2 \pi \ X}{\lambda} = \frac{2 \pi \times 10}{20} = \pi]$ is given by $\\$ $y_2 = a \ sin (wt - kx + \pi) \Rightarrow -a \ sin (wt-kx) = -1.5 mm$ $\\$ $\therefore displacement = -1.5mm$

15   A steel wire of length 64 cm weighs 5 g. If it is stretched by a force of 8 N, what would be the speed of a transverse wave passing on it ?

##### Solution :

mass = 5 g, length l =64 cm $\\$ $\therefore$ mass per unit length = $m = 5/64 \ g/cm$ $\\$ $\therefore$ Tension, $T=8N = 8 \times 10^{5} dync$ $\\$ $V=\sqrt{(T/m)} =\sqrt{(8 \times 10^{5} \times 64)/5} = 3200 cm/s = 32m/s$

16   A string of length 20 cm and linear mass density 0-40 g/cm is fixed at both ends and is kept under a tension of 16 N. A wave pulse is produced at t = 0 near an end as shown in figure (15-E3), which travels towards the other end. (a) When will the string have the shape shown in the figure again ? (b) Sketch the shape of the string at a time half of that found in part (a).

##### Solution :

A)Velocity of the wave,$v = \sqrt{(T/m)} = \sqrt{(16 \times 10^{5})/0.4} = 2000 cm/sec$ $\\$ $\therefore$ Time taken to reach to the other end = 20/2000 =0.01 sec $\\$ b)At t=0.01 s, there will be a 'though' at the right end as it is reflected.

17   A string of linear mass density 0'5 g/cm and a total length 30 cm is tied to a fixed wall at one end and to a frictionless ring at the other end (figure 15-E4). The ring can move on a vertical rod. A wave pulse is produced on the string which moves towards the ring at a speed of 20 cm/s. The pulse is symmetric about its maximum which is located at a distance of 20 cm from the end joined to the ring. (a) Assuming that the wave is reflected from the ends without loss of energy, find the time taken by the string to regain its shape. (b) The shape of the string changes periodically with time. Find this time period. (c) What is the tension in the string ?

##### Solution :

The crest reflects as a crest here, as the wire is travelling from denser to rarer medium. $\\$ $\Rightarrow$ phase change = 0 $\\$ a)To again original shape distance travelled by the wave S= 20+20 =40cm. $\\$ Wave speed, v=20m/s $\Rightarrow$ time =s/v = 40/20 = 2sec. $\\$ b)The wave regains its shape, after travelling a periodic distance = 2 \times 30 =60cm.$\\$ $\therefore$ Time period =60/20 =3sec. $\\$ c)Frequency, $n=(1/3 sec^{-1})$ $\\$ $n =(1/2I) \sqrt{(T/m)} \qquad$ m=mass per unit length = 0.5 g/cm $\\$ $\Rightarrow 1/3 = 1/(2 \times 30) \sqrt{(T/0.5)}$ $\\$ $\Rightarrow T= 400 \times 0.5 =200 dyne = 2 \times 10^{-3}$ Newton.

18   Two wires of different densities but same area of cross section are soldered together at one end and are stretched to a tension T. The velocity of a transverse wave in the first wire is double of that in the second wire. Find the ratio of the density of the first wire to that of the second wire.

##### Solution :

Let $v_1$= velocity in the $1^{st}$ string $\\$ $\Rightarrow v_1 =\sqrt{(T/m_1)}$ $\\$ Because $m_1$ = mass per unit length = $(\rho_1 a_1 l_1/l_1) = \rho_1a_1$ where $\\$ $a_1$=Area of cross section, $\\$ $\Rightarrow v_1 = \sqrt{(T/\rho_1a_1)} .....(1)$ $\\$ Let$v_2$ = velocity in the second string $\\$ $\Rightarrow v_2= \sqrt{(T/m^{2})}$ $\\$ $\Rightarrow v_2= \sqrt{(T/\rho_2a_2)} ..........(2)$ $\\$ Given that, $v_1 =2v_2$ $\\$ $\Rightarrow \sqrt{(T/\rho_1a_1)} = 2\sqrt{(T/\rho_2a_2)} \Rightarrow (T/a_1 \rho_1) = 4(T/a_2 \rho_2)$ $\\$ $\Rightarrow \rho_1/ \rho_2 = 1/4 \Rightarrow \rho_1 : \rho_2 = 1:4 \qquad$ (because a1 = a2)

19   A transverse wave described by $y = (0.02 m) \ sin \ [(1.0 m^{-1})x + (30 s^{-1})t]$ propagates on a stretched string having a linear mass density of $1.2 \times 10^{-4}$ kg/m. Find the tension in the string.

##### Solution :

m =mass per unit length = $1.2 \times 10^-{4}$ kg/mt $\\$ $y = (0.02 m) \ sin \ [(1.0 m^{-1})x + (30 s^{-1})t]$ $\\$ Here, $k=1 m^{-1} = 2 \pi / \lambda$ $\\$ $w= 30 s^{-1} = 2 \pi f$ $\\$ $\therefore$ velocity of the wave in the stretched string. $\\$ $v = \lambda f =w/k =30/1 =30 m/s$ $\\$ $\Rightarrow v = \sqrt{T/m} \Rightarrow 30\sqrt{(T/1.2) \times 10^{-4}N}$ $\\$ $\Rightarrow T=10.8 \times 10^{-2}N \Rightarrow T=1.08 \times 10^{-1}$ Newton.

20   A travelling wave is produced on a long horizontal string by vibrating an end up and down sinusoidally. The amplitude of vibration is 1.0 cm and the displacement becomes zero 200 times per second. The linear mass density of the string is 0'10 kg/m and it is kept under a tension of 90 N. (a) Find the speed and the wavelength of the wave. (b) Assume that the wave moves in the positive x-direction and at t = 0, the end x = 0 is at its positive extreme position. Write the wave equation. (c) Find the velocity and acceleration of the particle at x - 50 cm at time t = 10 ms.

##### Solution :

Amplitude, A=1cm,Tension T=90 N $\\$ Frequency, f=2000/2 =100Hz $\\$ mass per unit length, m=0.1 kg/mt $\\$ a)$\Rightarrow V= \sqrt{T/m} =30 m/s$ $\\$ $\lambda = V/f =30/100 = 0.3 m = 30cm$ $\\$ b)The wave equation $y =(1 cm) cos 2 \pi (t/0.01s) - (x/30 cm)$ $\\$ [because at x=0, displacement is maximum] $\\$ c) $y = 1 cos 2 \pi(x/30 -t/0.01)$ $\\$ $\Rightarrow v = dy/dt = (1/0.01)\ 2 \pi \ sin \ 2 \pi \big\{(x/30)- (t/0.01)\big\}$ $\\$ $\Rightarrow a= dv/dt = - \big\{ 4 \pi^{2}/(0.01)^{2}\big\} cos \ 2 \pi \big\{(x/30)- (t/0.01)\big\}$ $\\$ When,$x=50cm, t=10 ms =10 \times 10^{_3}$ $\\$ $x= (2 \pi /0.01) sin 2 \pi \big\{(5/3) - (0.01/0.01)\big\}$ $\\$ =$(p/0.01) sin (2 \pi \times 2/3) = (1/0.01) sin (4 \pi /3) = -200 \pi sin (\pi /3) = -200 \pi sin (\pi /3) = -200 \pi x (\sqrt 3/2)$ $\\$ =$544 cm/s =5.4 m/s$ $\\$ Similarly $\\$ $a = \big\{4 \pi^{2} /(0.01)^{2}\big\} cos 2 \pi \big\{(5/3) -1\big\}$ $\\$ = $4 \pi^{2} \times 10^{4} \times 1/2 \Rightarrow 2 \times 10^{5} cm/s^{2} \Rightarrow 2km/s^{2}$

21   A string of length 40 cm and weighing 10 g is attached to a spring at one end and to a fixed wall at the other end. The spring has a spring constant of 160 N/m and is stretched by 1.0 cm. If a wave pulse is produced on the string near the wall, how much time will it take to reach the spring ?

##### Solution :

$l =40 cm, mass = 10g$ $\\$ $\therefore$ mass per unit length, $m=10/40 =1/4 (g/cm)$ $\\$ spring constant $k= 160 N/m$ $\\$ deflection =$x=1 cm = 0.01 m$ $\\$ $\Rightarrow T=kx=160 \times 0.01 = 1.6N =16 \times 10^{4} dyne$ $\\$ Again $v =\sqrt{(T/m)} = \sqrt{(16 \times 10^4/(1/4)} = 8 \times 10^{2} cm/s =800 cm/s$ $\\$ $\therefore$ Time taken by the pulse to reach the spring $\\$ $t = 40/800 =1/20 = 0/05 sec$.

22   Two blocks each having a mass of 3.2 kg are connected by a wire CD and the system is suspended from the ceiling by another wire AB (figure 15-E5). The linear mass delity of the wire AB is 10 g/m and that of CD is 8 g/m. Find the speed of a transverse wave pulse produced in AB and in CD.

##### Solution :

$m_1 = m_2 =3.2kg$ $\\$ mass per unit length of AB= 10 g/mt =0.01 kg.mt $\\$ mass per unit length of CD = 8 g/mt = 0.008 kg/mt $\\$ for the string CD, T= 3.2 $\times$ g $\\$ $\Rightarrow v =\sqrt{(T/m)} = \sqrt{(3.2 \times 10)/0.008} = \sqrt{(32 \times 10^{3})/8} = 2 \times 10\sqrt{10} = 20 \times 3.14 = 63 m/s$ $\\$ for the string AB, $T= 2 \times 3.2 g =6.4 \times g =64N$ $\\$ $\Rightarrow v =\sqrt{(T/m)} =\sqrt{(64/0.01)} = \sqrt{6400} = 80 m/s$

23   In the arrangement shown in figure (15-E6), the string has a mass of 4.5 g. How much time will it take for a transverse disturbance produced at the floor to reach the pulley ? Take $g = 10 m/s^{2}$.

##### Solution :

Total length of string $2 + 0.25 = 2.25mt$ $\\$ Mass per unit length $m=\frac{4.5 \times 10^{-3}}{2.25} = 2 \times 10^{-3} kg/m$ $\\$ $T=2g=20 N$ $\\$ Wave speed, $v = \sqrt{(T/m)} = \sqrt{20}{\big(2 \times 10^{-3}\big)} = \sqrt{10^{4}} =10^{2} m/s =100m/s$ $\\$ Time taken to reach the pully, $t = (s/v) = 2/100 = 0.02 sec$.

24   A $4.0 \ kg$ block is suspended from the ceiling of an elevator through a, string having a linear mass density of $19.2 \times 10^{-3} \ k g/m$. Find the speed (with respect to the string) with which a wave pulse can proceed on the string if the elevator accelerates up at the rate of $2.0 m/s^2$. Take $g =l0 m/s^2$.

##### Solution :

$m = 19.2 \times 10^{-3} kg/m$ $\\$ from the freebody diagram, $\\$ $T - 4g -4a = 0$ $\\$ $\Rightarrow T= 4(a + g) = 48 \ N$ $\\$ Wave speed, $v = \sqrt{(T/m)} = 50 m/s$

25   A heavy ball is suspended from the ceiling of a motor car through a light string. A transverse pulse travels at a speed of 60 cm/s on the string when the car is at rest and 62 cm/s when the car accelerates on a horizontal road. Find the acceleration of the car. Take $g =10 \ m/s^{2}.$

##### Solution :

Let M = mass of the heavy ball $\\$ (m = mass per unit length) $\\$ Wave speed, $v_1 = \sqrt{(T/m)} = \sqrt{(Mg/m)}$ (because T= Mg) $\\$ $\Rightarrow 60 = \sqrt{(Mg/m)} \Rightarrow Mg/m = 60^{2} ....(1)$ $\\$ From the freebody diagram(2), $\\$ $v_2 = \sqrt{(T^{1}/m)}$ $\\$ $\Rightarrow v_2 = \frac{ \big[(Ma)^{2} + (Mg)^{2} \big]^\frac{1}{4}}{m^{1/2}} \qquad$ (because $T^{1} = \sqrt{(Ma^{2}) + (Mg^{2})}$ ) $\\$ $\Rightarrow 62 = \frac{ \big[(Ma)^{2} + (Mg)^{2} \big]^\frac{1}{4}}{m^{1/2}}$ $\\$ $\Rightarrow \frac{\sqrt{(Ma^{2}) + (Mg^{2})}}{m} = 62^{2} ....................(2)$ $\\$ $Eq (1) + Eq(2) \Rightarrow (Mg/m) \times \big [ m / \sqrt{(Ma^{2}) + (Mg^{2})} \big] = 3600/3844$ $\\$ $\Rightarrow g / \sqrt{(a^{2} + g^{2}}) = 0.936 \Rightarrow g^{2} / (a^{2} + g^{2}) = 0.876$ $\\$ $\Rightarrow (a^{2} + 100) 0.876 =100$ $\\$ $\Rightarrow a^{2} \times 0.876 =100 - 87.6 = 12.4$ $\\$ $\Rightarrow a^{2} = 12.4/0.876 = 14.15$ $\Rightarrow a = 3.76 \ m/s^{2}$ $\\$ $\therefore$ Acce of the car =$3.7 \ m/s^{2}$

26   A circular loop of string rotates about its axis on a frictionless horizontal plane at a uniform rate so that the tangential speed of any particle of the string is v. If a small transverse disturbance is produced at. a point of the loop, with what speed (relative to the string) will this disturbance travel on the string ?

##### Solution :

m = mass per unit length of string $\\$ R = Radius of the loop $\\$ w = angular velocity, V =linear velocity of the string $\\$ Consider one half of the string as shown in figure. $\\$ the half loop experiences cetrifugal force at evary point, away from centre, which is balanced by tension 2T. $\\$ Consider an element of angular part $d\theta$ at angle $\theta$. Consider another element symmetric to this centrifugal force experienced by the element = $(mRd \theta)w^{2}R$. $\\$ (..Length of element = $Rd\theta, mass =mRd\theta$)$\\$ Resolving into rectangular components net force on the two symmetric elements, $\\$ $DF = 2MR^{2} d \theta w^{2} \ sin \theta$ [horizontal components cancels each other] $\\$ So, total $F = \int_{0}^{\pi /2} 2mR^{2} w^{2} \ sin \theta d \theta = 2mR^{2}w^{2} [-cos \theta] \Rightarrow 2mR^{2}w^{2}$ $\\$ Again velocity, $2T = 2mR^{2} w^{2} \qquad \Rightarrow T = mR^{2} w^{2}$ $\\$ Velocity of transverse vibration $V = \sqrt{T/m} = wR = V$ $\\$ So, the speed of disturbance will be V.

27   A heavy but uniform rope of length L is suspended from a ceiling. (a) Write the velocity of a transverse wave travelling on the string as a function of the distance from the lower end. (b) If the rope is given a sudden sideways jerk at the bottom, how long will it take for the pulse to reach the ceiling ? (c) A particle is dropped from the ceiling at the instant the bottom end is given the jerk. Where will the particle meet the pulse ?

##### Solution :

a) m $\rightarrow$ mass per unit of length of string consider an element at distance ‘x’ from lower end. Here wt acting down ward = (mx)g = Tension in the string of upper part $\\$ Velocity of transverse vibration =$v = \sqrt{T /m} = \sqrt{(mgx /m)} = \sqrt {(gx)}$ $\\$ b) For small displacement $dx, dt = dx / \sqrt{(gx)}$ Total time $T = \int_{0}^{L} dx / \sqrt{gx} = \sqrt{(4L/g)}$ $\\$ c) Suppose after time ‘t’ from start the pulse meet the particle at distance y from lower end. $\\$ $t = \int_{0}^{L} dx / \sqrt{gx} = \sqrt{(4y/g)}$ $\\$ $\therefore$ Distance travelled by the particle in this time is (L – y) $\\$ $\therefore S - ut + 1/2 gt^2$ $\\$ $\Rightarrow L - y(1/2)g \times \big\{ \sqrt{ (4y / g)^{2}} \big\}$ $\qquad \big\{u=0 \big\}$ $\\$ $\Rightarrow L - y = 2y \Rightarrow 3y = L$ $\\$ $\Rightarrow y = L/3$.So,the particle meet at distance L/3 from lower end.

28   Two long strings A and B, each having linear mass density 1.2 x 10 2 kg/m, are stretched by different tensions 4.8 N and 7.5 N respectively and are kept parallel to each other with their left ends at x = 0. Wave pulses are produced on the strings at the left ends at t = 0 on string A and at t = 20 ms on string B. When and where will the pulse on B overtake that on A ?

##### Solution :

$m_A = 1.2 \times 10^{-2} kg/m, T_A = 4.8 \ N$ $\\$ $\Rightarrow V_A = \sqrt{T/m} =20m/s$ $\\$ $m_B = 1.2 \times 10^{-2} kg/m, T_B= 7.5 \ N$ $\\$ $\Rightarrow V_B = \sqrt{T/m} =25m/s$ $\\$ t = 0 in string A $\\$ t1 = 0 + 20 ms = $20 \times 10^{-3}$ = 0.02 sec $\\$ In 0.02 sec A has travelled $20 \times 0.02 = 0.4 mt$ $\\$ Relative speed between A and B = 25 – 20 = 5 m/s $\\$ Time taken for B for overtake A = s/v = 0.4/5 = 0.08 sec

29   A transverse wave of amplitude 0.50 mm and frequency 100 Hz is produced on a wire stretched to a tension of 100 N. If the wave speed is 100 m/s, what average power is the source transmitting to the wire ?

##### Solution :

$r =0.5mm = 0.5 \times 10^{-3}mt$ $\\$ $f =100 \ Hz, T=100 \ N$ $\\$ $V =100 m/s$ $\\$ $v = \sqrt{T/m} \Rightarrow v^{2} = (T/m) \Rightarrow m = (T/v^{2}) = 0.01 kg/m$ $\\$ $P_{ave} = 2 \pi^{2} mvr^{2}f^{2}$ $\\$ $=2(3.14)^{2}(0.01) \times 100 \times (0.5 \times 10^{-3})^{2} \times (100)^{2} \Rightarrow 49 \times 10^{-3}watt =49 \ mW$

30   A 200 Hz wave with amplitude 1 mm travels on a long string of linear mass density 6 g/m kept under a tension of 60 N. (a) Find the average power transmitted across a given point on the string. (b) Find the total energy associated with the wave in a 2.0 m long portion of the sring.

##### Solution :

$A = 1 mm = 10^{–3} m, \ m = 6 g/m = 6 \times 10^{-3} kg/m$ $\\$ $T = 60 N, f = 200 Hz$ $\\$ $\therefore V = \sqrt{T /m} = 100 m/s$ $\\$ a) $P_{average} = 2 \pi^{2} \ mv A^{2}f^{2} = 0.47 \ W$ $\\$ b) Length of the string is 2 m. So, t = 2/100 = 0.02 sec. $\\$ $Energy = 2\pi^{2} \ mvf^{2}A^{2}t = 9.46 \ mJ$.

31   A tuning fork of frequency 440 Hz is attached to a long string of linear mass density 0.01 kg/m kept under a tension of 49 N. The fork produces transverse waves of amplitude 0.50 mm on the string. (a) Find the wave speed and the wavelength of the waves. (b) Find the maximum speed and acceleration of a particle of the string. (c) At what average rate is the tuning fork transmitting energy to the string ?

##### Solution :

$f = 440 Hz, m = 0.01 \ kg/m, \ T = 49 \ N, r = 0.5 \times 10^{–3} \ m$ $\\$ a)$v = \sqrt {T /m} = 70 \ m/s$ $\\$ b) $v = \lambda f \Rightarrow \lambda = v/f = 16 \ cm$ $\\$ c) $P_{average} = 2 \pi^{2} \ mvr^{2}f^{2} = 0.67 \ W.$

32   Two waves, travelling in the same direction through the same region, have equal frequencies, wavelengths and amplitudes. If the amplitude of each wave is 4 mm and the phase difference between the waves is 90°, what is the resultant amplitude ?

##### Solution :

Phase difference $\phi = \pi /2$ $\\$ f and $\lambda$ are same. So, $\omega$ is same. $\\$ $y_1 = r \ sin wt, y_2 = rsin(wt + \pi /2)$ $\\$ From the principle of superposition $\\$ $y = y1 + y2 \rightarrow \qquad = r \ sin wt + r \ sin(wt + \pi /2)$ $\\$ $\qquad \ = r[ sin wt + sin(wt + \pi /2]$ $\\$ $\qquad \ =r[2sin \big\{(wt + wt + \pi/2)/2 \big\} \ cos \big\{(wt - wt - \pi/2)/2 \big\}]$ $\\$ $\Rightarrow y= 2r \ sin(wt + \pi /4) cos(- \pi/4)$ $\\$ Resultant amplitude = $\sqrt{2} \ r = 4 \sqrt{2} mm \qquad (because r = 4 mm)$

33   Figure (15-E7) shows two wave pulses at t = 0 travelling on a string in opposite directions with the same wave speed 50 cm/s. Sketch the shape of the string at t = 4 ms, 6 ms, 8 ms, and 12 ms.

##### Solution :

The distance travelled by the pulses are shown below. $\\$ $t = 4 ms = 4 \times 10^{–3} s \qquad s = vt = 50 \times 10 \times 4 \times 10^{–3} = 2 \ mm$ $\\$ $t = 8 ms = 8 \times 10^{–3} s \qquad s = vt = 50 \times 10 \times 8 \times 10^{–3} = 4 \ mm$ $\\$ $t = 6 ms = 6 \times 10^{–3} s \qquad s = 3 \ mm$ $\\$ $t = 12 ms = 12 \times 10^{–3} s \qquad s = 50 \times 10 \times 12 \times 10^{–3} = 6 \ mm$ $\\$ The shape of the string at different times are shown in the figure.

34   Two waves, each having a frequency of 100 Hz and a wavelength of 2.0 cm, are travelling in the same direction on a string. What is the phase difference between the waves (a) if the second wave was produced 0.015 s later than the first one at the same place, (b) if the two waves were produced at the same instant but the first one was produced a distance 4.0 cm behind the second one?(c) If each of the waves has an amplitude of 2.0 mm, what would be the amplitudes of the resultant waves in part (a) and (b)? ##### Solution :

$f = 100 Hz, \lambda = 2 cm = 2 \times 10^{–2} m$ $\\$ $\therefore$ wave speed, $v = f \lambda = 2 \ m/s$ $\\$ a) in 0.015 sec 1st wave has travelled $x = 0.015 \times 2 = 0.03 \ m$ = path $diff^{n}$ $\\$ $\therefore$ corresponding phase difference, $\phi = 2 \pi X/ \lambda = \big\{ 2 \pi / (2 \times 10^{-2})\big\} \times 0.03 = 3 \pi.$ $\\$ b) Path different $x = 4 cm = 0.04 m$ $\\$ $\Rightarrow \phi = (2 \pi / \lambda)X = \big\{(2 \pi /2 \times 10^{-2}) \times 0.04 \big\} = 4 \pi.$ $\\$ c) The waves have same frequency, same wavelength and same amplitude. $\\$ Let, $y1 = r sin wt, y2 = r sin (wt + \phi)$ $\\$ $\Rightarrow y = y1 + y2 = r[sin wt + (wt + \phi)]$ $\\$ $= 2r \ sin (wt + \phi/2) cos (\phi/2)$ $\\$ $\therefore$ resultant amplitude = $2r \ cos \phi/2$ $\\$ So, when $\phi = 3 \pi, r = 2 \times 10^{–3} m$ $\\$ $R_{res} = 2 \times (2 \times 10^{–3}) \ cos (3 \pi /2) = 0$ $\\$ Again, when $\phi = 4 \pi, R_{res} = 2 \times (2 \times 10^{–3}) \ cos (4 \pi /2) = 4 mm.$

35   If the speed of a transverse wave on a stretched string of length 1 m is 60 m/s, what is the fundamental frequency of vibration ?

##### Solution :

$l = 1 m, V = 60 m/s$ $\\$ $\therefore$ fundamental frequency, $f_{0} = V/2l = 30 sec^{–1} = 30 Hz.$

36   A wire of length 2.00 m is stretched to a tension of 160 N. If the fundamental frequency of vibration is 100 Hz, find its linear mass density.

##### Solution :

$l = 2m, f0 = 100 Hz, T = 160 \ N$ $\\$ $f_0 = 1/ 2l \sqrt{(T /m)}$ $\\$ $\Rightarrow m = 1 \ g/m.$ So, the linear mass density is $1 \ g/m.$

37   A steel wire of mass 4.0 g and length 80 cm is fixed at the two ends. The tension in the wire is 50 N. Find the frequency and wavelength of the fourth harmonic of the fundamental.

##### Solution :

$m = (4/80) g/ cm = 0.005 kg/m$ $\\$ $T = 50 N, l = 80 cm = 0.8 m$ $\\$ $v = \sqrt {(T /m)} = 100 m/s$ $\\$ fundamental frequency $f_0 = 1/ 2l \sqrt{ (T /m)} = 62.5 Hz$ $\\$ First harmonic = 62.5 Hz $\\$ f4 = frequency of fourth harmonic $= 4f_0 = F_3 = 250 \ Hz$ $\\$ $V = f_4 \lambda_4 \Rightarrow \lambda_4 = (v/f4) = 40 cm.$

38   A piano wire weighing , 6.00 g and having a length of 90.0 cm emits a fundamental frequency corresponding to the "Middle C" (v - 261.63 Hz). Find the tension in the wire.

##### Solution :

$l = 90 \ cm = 0.9 \ m$ $\\$ $m = (6/90) \ g/cm = (6/900) \ kg/mt$ $\\$ $f = 261.63 Hz$ $\\$ $f = 1/ 2l (T /m) \Rightarrow T = 1478.52 \ N = 1480 \ N.$

39   A sonometer wire having a length of 1.50m between the bridges vibrates in its second harmonic in resonance with a tuning fork of frequency 256 Hz. What is the speed of the transverse wave on the wire ?

##### Solution :

First harmonic be $f_0$, second harmonic be $f_1$ $\\$ $\therefore f_1 = 2f_0$ $\\$ $\Rightarrow f_0 = f_1/2$ $\\$ $f_1 = 256 Hz$ $\\$ $\therefore$ 1st harmonic or fundamental frequency $\\$ $f_0 = f_1/2 = 256 / 2 = 128 Hz$ $\\$ $\lambda/2 = 1.5 m \Rightarrow \lambda= 3m$ (when fundamental wave is produced) $\\$ $\Rightarrow Wave speed = V = f_0Ql = 384 \ m/s.$

40   The length of the wire shown in figure (15-E8) between the pulleys is 1.5 m and its mass is 12.0 g. Find the frequency of vibration with which the wire vibrates in two loops leaving the middle point of the wire between the pulleys at rest.

##### Solution :

$l = 1.5 m, mass – 12 g$ $\\$ $\Rightarrow m = 12/1.5 g/m = 8 \times 10^{–3} kg/m$ $\\$ $T = 9 \times g = 90 \ N$ $\lambda = 1.5 m, f_1 = 2/2l \sqrt{T /m}$ $\\$ [for, second harmonic two loops are produced] $\\$ $f_1 = 2f_0 \Rightarrow 70 \ Hz.$

41   A one metre long stretched string having a mass of 40 g is attached to a tuning fork. The fork vibrates at 128 Hz in a direction perpendicular to the string. What should be the tension in the string if it is to vibrate in four loops ?

##### Solution :

A string of mass 40 g is attached to the tuning fork $\\$ $m = (40 \times 10^{–3}) kg/m$ $\\$ The fork vibrates with f = 128 Hz $\\$ $\lambda = 0.5 \ m$ $\\$ $v = f \lambda = 128 \times 0.5 = 64 m/s$ $\\$ $v = \sqrt{ T /m} \Rightarrow T = v^{2}m = 163.84 \ N \Rightarrow 164 \ N.$

42   A wire, fixed at both ends is seen to vibrate at a resonant frequency of 240 Hz and also at 320 Hz. (a) What could be the maximum value of the fundamental frequency ? (b) If transverse waves can travel on this string at a speed of 40 m/s, what is its length ?

##### Solution :

This wire makes a resonant frequency of 240 Hz and 320 Hz. $\\$ The fundamental frequency of the wire must be divisible by both 240 Hz and 320 Hz. $\\$ a) So, the maximum value of fundamental frequency is 80 Hz. $\\$ b) Wave speed, v = 40 m/s $\\$ $\Rightarrow 80 = (1/2l) \times 40 \Rightarrow 0.25 m.$

43   A string, fixed at both ends, vibrates in a resonant mode with a separation of 2.0 cm between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to 1.6 cm. Find the length of the string.

##### Solution :

Let there be ‘n’ loops in the 1st case $\\$ $\Rightarrow$ length of the wire, l = $(n \lambda_1)/2 \qquad [\lambda_1 = 2 \times 2 =4 cm]$ $\\$ So there are (n+1)loops with the $2^nd$ case $\\$ $\Rightarrow$ length of the wire,$l = \big\{(n +1) \lambda_{2}/2 \qquad [\lambda = 2 \times 1.6 =3.2cm]$ $\\$ $\Rightarrow n \lambda_{1} / 2 = \frac{(n+1)\lambda_{2}}{2}$ $\\$ $\Rightarrow n \times 4 = (n + 1) (3.2) \Rightarrow n = 4$ $\\$ $\therefore$ length of the string, $l = (n \lambda_{1})/2 =8cm$

44   A 660 Hz tuning fork sets up vibration in a string clamped at both ends. The wave speed for a transverse wave on this string -is 220 m/s and the string vibrates in three loops. (a) Find the length of the string. (b) If the maximum amplitude of a particle is 0.5 cm, write a suitable equation describing the motion.

##### Solution :

Frequency of the tuning fork, f = 660 Hz $\\$ Wave speed, $v = 220 m/s \Rightarrow \lambda = v/f = 1/3 m$ $\\$ No.of loops = 3 $\\$ a) So,$f = (3/2l)v \Rightarrow l = 50 cm$ $\\$ b) The equation of resultant stationary wave is given by $\\$ $y = 2A \ cos (2 \pi x/Ql) \ sin (2 \pi vt/ \lambda)$ $\\$ $\Rightarrow y = (0.5 cm) \ cos (0.06 \pi \ cm^{–1}) sin (1320 \pi s^{–1}t)$

45   A particular guitar wire is 30.0 cm long and vibrates at a frequency of 196 Hz when no finger is placed on it. The next higher notes on the scale are 220 Hz, 247 Hz, 262 Hz and 294 Hz. How far from the end of the string must the finger be placed to play these notes ?

##### Solution :

l1 = 30 cm = 0.3 m $\\$ f1 = 196 Hz, f2 = 220 Hz $\\$ We know f $\propto$ (1/l) (as V is constant for a medium)$\\$ $\Rightarrow \frac{f_1}{f_2} = \frac{l_2}{l_1} \Rightarrow l_2 = 26.7cm$ $\\$ Again f3 = 247 Hz $\\$ $\Rightarrow = \frac{f_3}{f_1} = \frac{l_1}{l_3} \Rightarrow \frac{0.3}{l_3}$ $\\$ $\Rightarrow l3 = 0.224 m = 22.4 cm and l3 = 20 cm$

46   A steel wire fixed at both ends has a fundamental frequency of 200 Hz. A person can hear sound of maximum frequency 14 kHz. What is the highest harmonic that can be played on this string which is audible to the person ?

##### Solution :

Fundamental frequency f1 = 200 Hz $\\$ Let $l_4$ Hz be nth harmonic $\\$ $\Rightarrow F_2/F_1 = 14000/200$ $\\$ $\Rightarrow NF_1/F_1 = 70 \Rightarrow N = 70$ $\\$ $\Rightarrow$ The highest harmonic audible is $70^{th}$ harmonic.

47   Three resonant frequencies of a string are 90, 150 and 210 Hz. (a) Find the highest possible fundamental frequency of vibration of this string. (b) Which harmonics of the fundamental are the given frequencies ? (c) Which overtones are these frequencies. (d) If the length of the string is 80 cm, what would be the speed of a transverse wave on this string ?

##### Solution :

The resonant frequencies of a string are $\\$ f1 = 90 Hz, f2 = 150 Hz, f3 = 120 Hz $\\$ a) The highest possible fundamental frequency of the string is f = 30 Hz $\\$ [because f1, f2 and f3 are integral multiple of 30 Hz] $\\$ b) The frequencies are f1 = 3f, f2 = 5f, f3 = 7f $\\$ So, f1, f2 and f3 are 3rd harmonic, 5th harmonic and 7th harmonic respectively. $\\$ c) The frequencies in the string are f, 2f, 3f, 4f, 5f, ………. $\\$ So, 3f = 2nd overtone and 3rd harmonic $\\$ 5f = 4th overtone and 5th harmonic $\\$ 7f = 6th overtone and 7th harmonic $\\$ d) length of the string is l = 80 cm $\\$ $\Rightarrow$ f1 = (3/2l)v (v = velocity of the wave) $\\$ $\Rightarrow 90 = \big\{3/(2 \times 80)\big\} \times K$ $\\$ $\Rightarrow K = (90 \times 2 \times 80) / 3 = 4800 cm/s = 48 m/s.$

48   Two wires are kept tight between the same pair of supports. The tensions in the wires are in the ratio 2 : 1, the radii are in the ratio 3 : 1 and the densities are in the ratio 1 :2. Find the ratio of their fundamental frequencies.

##### Solution :

Frequency $f = \frac{1}{ID} \sqrt{ \frac{T}{ \pi \rho}} \Rightarrow f_1 = \frac{1}{I_1D_1} \sqrt{ \frac{T_1}{ \pi \rho_1}} \Rightarrow f_2 = \frac{1}{I_2n_2} \sqrt{ \frac{T_2}{ \pi \rho_2}}$ $\\$ Given that, T1/T2 = 2, r1 / r2 = 3 = D1/D2 $\\$ $\frac{\rho_1}{\rho_2} = \frac{1}{2}$ $\\$ So $\frac{f_1}{f_2} = \frac{I_2D_2}{I_1D_1} \sqrt{\frac{T_1}{T_2}} \sqrt{\frac{\pi \rho_2}{\pi \rho_1}} \qquad (I_1 = I_2 = length of string)$ $\\$ $\Rightarrow f_1 : f_2 = 2 : 3$

49   A uniform horizontal rod of length 40 cm and mass 1.2 kg is supported by two identical wires as shown in figure (15-E9). Where should a mass of 4.8 kg be placed on the rod so that the same tuning fork may excite the wire on left into its fundamental vibrations and that on right into its first overtone ? Take g = 10 m/s 2.

##### Solution :

Length of the rod = L = 40 cm = 0.4 m $\\$ Mass of the rod m = 1.2 kg $\\$ Let the 4.8 kg mass be placed at a distance ‘x’ from the left end. $\\$ Given that, fl = 2fr $\\$ $\therefore \frac{1}{2l} \sqrt{\frac{T_1}{m}} = \frac{2}{2l} \sqrt{\frac{T_r}{m}}$ $\\$ $\Rightarrow \sqrt{\frac{T_I}{T_r}} = 2 \Rightarrow \frac{T_1}{T_r} =4 ..........(1)$ $\\$ From the freebody diagram, $\\$ $T_l + T_r = 60 N$ $\\$ $\Rightarrow 4T_r +T_r = 60 N$ $\\$ $\therefore T_r = 12 N$ and $T_l = 48 N$ $\\$ Now taking moment about point A, $T_r \times (0.4) = 48x + 12 (0.2) \Rightarrow x = 5 cm$ $\\$ So, the mass should be placed at a distance 5 cm from the left end.

50   Figure (15-E10) shows an aluminium wire of length 60 cm joined to a steel wire of length 80 cm and stretched between two fixed supports. The tension produced is 40 N. The cross-sectional area of the steel wire is $1.0 mm^2$and that of the aluminium wire is $3.0 mm^2$. What could be the minimum frequency of a tuning fork which can produce standing waves in the system with the joint as a node ? The density of aluminium is $2.6 g/cm^3$ and that of steel is $7.8 g/cm^3$.

##### Solution :

51   Figure (15-E10) shows an aluminium wire of length 60 cm joined to a steel wire of length 80 cm and stretched between two fixed supports. The tension produced is 40 N. The cross-sectional area of the steel wire is $1.0 mm^2$and that of the aluminium wire is $3.0 mm^2$. What could be the minimum frequency of a tuning fork which can produce standing waves in the system with the joint as a node ? The density of aluminium is $2.6 g/cm^3$ and that of steel is $7.8 g/cm^3$.

##### Solution :

$\rho_s = 7.8 \ g/cm^3 , \rho_A = 2.6 \ g/cm^3$ $\\$ $m_s = \rho_s A_s = 7.8 \times 10^{–2} \ g/cm \qquad$ (m = mass per unit length) $\\$ $mA = \rho_A A_A = 2.6 \times 10^{–2} \times 3 \ g/cm = 7.8 \times 10^{–3} \ kg/m$ $\\$ A node is always placed in the joint. Since aluminium and steel rod has same mass per unit length, velocity of wave in both of them is same. $\\$ $\Rightarrow v = \sqrt{T /m} \Rightarrow 500/7 \ m/x$ $\\$ For minimum frequency there would be maximum wavelength for maximum wavelength minimum no of loops are to be produced.$\\$ $\therefore$ maximum distance of a loop = 20 cm $\\$ $\Rightarrow$ wavelength = $\lambda = 2 \times 20 = 40 cm = 0.4 m$ $\\$ $\therefore$ $f$ = v/$\lambda = 180 Hz.$

52   A string of length L fixed at both ends vibrates in its fundamental mode at a frequency v and a maximum amplitude A. (a) Find the wavelength and the wave number k. (b) Take the origin at one end of the string and the X-axis along the string. Take the Y-axis along the direction of the displacement. Take t = 0 at the instant when the middle point of the string passes through its mean position and is going towards the positive y-directian. Write the equation describing the standing wave:

##### Solution :

Fundamental frequency $\\$ $V = 1/2l \sqrt{T /m} \Rightarrow T /m = v2l \qquad$ [ T /m = velocity of wave] $\\$ a) wavelength, $\lambda$ = velocity / frequency = v2l / v = 2l and wave number =$K = 2 \pi /\lambda = 2 \pi/2l = \pi/l$ $\\$ b) Therefore, equation of the stationary wave is $\\$ $y = A \ cos (2\pi x/\lambda) \ sin (2 \pi Vt / L)$ $\\$ =$A \ cos (2\pi x / 2l) \ sin (2 \pi Vt / 2L)$ $\\$ $v = V/2L \qquad [because v = (v/2l)]$

53   A 2 in long string fixed at both ends is set into vibrations in its first overtone. The wave speed on the string is 200 m/s and thee amplitude is 0.5 cm. (a) Find the wavelength and the frequency. (b) Write the equation giving the displacement of different points as a function of time. Choose the X-axis along the string with the origin at one end and t = 0 at the instant when the point x = 50 cm has reached its maximum displacement.

##### Solution :

$V = 200 m/s, 2A = 0.5 m$ $\\$ a) The string is vibrating in its $1^{st}$ overtone $\\$ $\Rightarrow \lambda = 1 = 2m$ $\\$ $\Rightarrow f = v/\lambda = 100 Hz$ $\\$ b) The stationary wave equation is given by $\\$ $y = 2A \ cos \frac{2 \pi X}{\lambda} \ sin \frac{2 \pi V t}{\lambda}$ $\\$ = $(0.5 cm) cos [(\pi m^{–1})x] sin [(200 \pi s^{–1})t]$

54   The equation for the vibration of a string, fixed at both ends vibrating in its third harmonic, is given by $y = (0.4 cm) sin[(0.314 cm^{-1}) x] cos[(600 \pi s^{-1})t]$ . (a) What is the frequency of vibration ? (b) What are the positions of the nodes ? (c) What is the length of the string ? (d) What is the wavelength and the speed of two travelling waves that can interfere to give this vibration ?

##### Solution :

The stationary wave equation is given by $\\$ $y = (0.4 cm) sin [(0.314 cm^{– 1})x] cos [(6.00 \pi s^{–1})t]$ $\\$ a) $\omega = 600 \pi \Rightarrow 2 \pi f = 600 \pi \Rightarrow f = 300 Hz$ $\\$ wavelength, $\lambda = 2 \pi /0.314 = (2 \times 3.14) / 0.314 = 20 cm$ $\\$ b) Therefore nodes are located at, 0, 10 cm, 20 cm, 30 cm $\\$ c) Length of the string = $3\lambda/2 = 3 \times 20/2 = 30 cm$ $\\$ d) $y = 0.4 sin (0.314 x) cos (600 \pi t) \Rightarrow 0.4 sin \big\{(\pi/10)x\big\} cos (600 \pi t)$ $\\$ since, $\lambda$ and v are the wavelength and velocity of the waves that interfere to give this vibration $\lambda = 20cm$ $\\$ $v = \omega/k = 6000 cm/sec =60m/s$

55   The equation of a standing wave, produced on a string fixed at both ends, is $y = (0.4 cm) sin[(0.314 cm^{-1}) x] cos[(600 \pi s^{-1})t]$ What could be the smallest length of the string ?

##### Solution :

The equation of the standing wave is given by $\\$ $y = (0.4 cm) sin[(0.314 cm^{-1}) x] cos[(600 \pi s^{-1})t]$ $\\$ $\Rightarrow k = 0.314 = \pi/10$ $\\$ $\Rightarrow 2\pi/ \lambda = \pi/10 \Rightarrow \lambda= 20 cm$ $\\$ for smallest length of the string, as wavelength remains constant, the string should vibrate in fundamental frequency $\\$ $\Rightarrow l = \lambda/2 = 20 cm / 2 = 10 cm$

56   A 40 cm wire having a mass of 3.2 g is stretched between two fixed supports 40.05 cm apart. In its fundamental mode, the wire vibrates at 220 Hz. If the area of cross-section of the wire is $1.0 mm^{2}$, find its Young's modulus.

##### Solution :

$L = 40 cm = 0.4 m, mass = 3.2 kg = 3.2 \times 10^{–3} kg$ $\\$ $\therefore$ mass per unit length, $m = (3.2)/(0.4) = 8 \times 10^{–3} kg/m$ $\\$ change in length, $\Delta L = 40.05 – 40 = 0.05 \times 10^{–2} m$ $\\$ strain =$\Delta L/L = 0.125 \times 10^{–2} m$ $\\$ $f = 220 Hz$ $\\$ $f = \frac{1}{2l} \sqrt{\frac{T}{m}}$ Strain = $248.19/1 mm^{2} = 248.19 \times 106$ $\\$ $Y = stress / strain = 1.985 \times 10^{11} N/m^{2}.$

57   Figure (15-E11) shows a string stretched by a block going over a pulley . The string vibrates in its tenth harmonic in unison with a particular tuning fork. When a beaker containing water is brought under the block so that the block is completely dipped into the beaker, the string vibrates in its eleventh harmonic. Find the density of the material of the block.

##### Solution :

Let, $\rho \rightarrow$ density of the block $\\$ Weight $\rho$ Vg where V = volume of block $\\$ The same turning fork resonates with the string in the two cases $\\$ $f_10 = \frac{10}{2l} \sqrt{ \frac{T - \rho_w V_g}{m}} = \frac{11}{2l} \sqrt{ \frac{(\rho - \rho_w)V_g}{m}}$ $\\$ As the f of tuning fork is same, $\\$ $f_{10} = f_{11} \Rightarrow \frac{10}{2l} \sqrt{ \frac{\rho V_g}{m}} = \frac{11}{2l} \sqrt{ \frac{\rho - \rho_w)V_g}{m} }$ $\\$ $\Rightarrow \frac{10}{11} = \sqrt { \frac{\rho - \rho_w}{m} } \Rightarrow \frac{\rho -1}{\rho} = \frac{100}{121} \qquad$ $(because, \rho_w= 1 \ gm/cc)$ $\\$ $\Rightarrow 100 \rho = 121 \rho – 121 \Rightarrow 5.8 \times 10^3 kg/m^3$

58   A 2.00 m long rope, having a mass of 80 g, is fixed at one end and is tied to a light string at the other end. The tension in the string is 256 N. (a) Find the frequencies of the fundamental and the first two overtones. (b) Find the wavelength in the fundamental and the first two overtones.

##### Solution :

l = length of rope = 2 m $\\$ M = mass = 80 gm = 0.8 kg $\\$ mass per unit length = m = 0.08/2 = 0.04 kg/m Tension T = 256 N $\\$ Velocity, $V = \sqrt {T /m} = 80 m/s$ $\\$ For fundamental frequency, $\\$ $l = \lambda/4 \Rightarrow \lambda = 4l = 8 m$ $\\$ $\Rightarrow f = 80/8 = 10 Hz$ $\\$ a) Therefore, the frequency of 1st two overtones are $\\$ 1st overtone = 3f = 30 Hz $\\$ 2nd overtone = 5f = 50 Hz $\\$ b) $\lambda_11 = 4l = 8 m$ $\\$ $\lambda_1 = V/ f1 = 2.67 m$ $\\$ $\lambda_2 = V/f2 = 1.6 mt$

59   A heavy string is tied at one end to a movable support and to a light thread at the other end as shown in figure (15-E12). The thread goes over a fixed pulley and supports a weight to produce a tension. The lowest frequency with which the heavy string resonates is 120 Hz. If the movable support is pushed to the right by 10 cm so that the joint is placed on the pulley, what will be the minimum frequency at which the heavy string can resonate ?

##### Solution :

Initially because the end A is free, an antinode will be formed.$\\$ $So, l = Ql1 / 4$ $\\$ Again, if the movable support is pushed to right by 10 m, so that the joint is placed on the pulley, a node will be formed there.$\\$ So,$l = \lambda_2 / 2$ $\\$ Since, the tension remains same in both the cases, velocity remains same. $\\$ As the wavelength is reduced by half, the frequency will become twice as that of 120 Hz i.e. 240 Hz.