Waves Test

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Waves Test - 67

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Weekly Quiz Competition

Question 1
1 / -0

Two sine waves travel in the same direction in a medium. The amplitude of each wave is A and the phase difference between the two waves is $$120^{\circ}$$. The resultant amplitude will be

A
$$A$$

B
$$2A$$

C
$$4A$$

D
$$\sqrt{2 A}$$

Solution
Question 2
1 / -0

Two point lie on a ray are emerging from a source of simple harmonic wave having period $$0.045$$. The wave speed is $$300\ m/s$$ and points are at $$10\ m$$ and $$16\ m$$ from the source. They differ in phase by :

A
$$\pi$$

B
$$\pi/2$$

C
$$0$$ or $$2\pi$$

D
$$none\ of\ these$$

Solution
Question 3
1 / -0

A simple harmonic wavetrain of amplitude $$5\ cm$$ and frequency $$100\ Hz$$ is travelling in the positive $$x-$$ direction with a velocity of $$30\ m/s$$. The displacement velocity and acceleration at $$t=3s$$ of a particle of the medium situated $$100\ cm$$ from the origin are respectively.

A
$$-3.44\ cm+1750\ cm/s, 17\times cm/s^{2}$$

B
$$4.33\ cm+1570\ cm/s,71\times 10^{4}\ cm/s^{2}$$

C
$$-4.33\ cm,+1570\ cm/s,171\times 10^{4}\ cm/s^{2}$$

D
$$-3.44\ cm,-1750\ cm/s,171\times 10^{4}\ cm/s^{2}$$

Solution

$$y=A\sin (\dfrac{2\pi \mu}{v}(vt-x))$$

$$y=5\sin (\dfrac{2\pi 100}{3000}(3000\times 3-100))=-4.45cm$$

velocity= $$\dfrac{dy}{dx}=A\times 2\pi\mu\cos(\dfrac{2\pi\mu}{v}(vt-x))=5\times 2\pi\times 100\cos(\dfrac{2\pi\times 100}{3000}(3000\times 3-180))=1571cm/s$$

accleration=$$\dfrac{d^2y}{dx^2}=-A\times 4\pi^2\mu^2sin(\dfrac{2\pi\mu}{v}(vt-x))=-5\times 4\pi^2\times 10000\sin(\dfrac{2\pi\times 100}{3000}(3000\times 3-180))=171\times 10^{4}cm/s^2$$
Question 4
1 / -0

Two identical pulses move in opposite directions with same uniform speeds on a stretched string. The width and kinetic energy of each pulse is $$L$$ and $$k$$ respectively. At the instant they completely overlap, the kinetic energy of the width $$L$$ of the string where they overlap is:

A
$$k$$

B
$$2k$$

C
$$4k$$

D
$$8k$$

Solution

$$y_1=a\sin(2\pi x /L)$$
When they get combined: $$y_2=2asin(2\pi x/L)$$
We know particle velocity= wave velocity x dy/dx
As in both the cases wave velocities are the same
$$\dfrac{v_1}{v_2}=\dfrac{\dfrac{dy_1}{dx}}{\dfrac{dy_2}{dx}}=\dfrac{1}{2}\Rightarrow v_2=2v_1$$

$$\dfrac{KE_1}{KE_2}=\dfrac{{v_1}^2}{{v_2}^2}=\dfrac{1}{4}\Rightarrow KE_2=4\times KE_1=4k$$
Question 5
1 / -0

A wave motion has the function $$Y=a_{0}\sin (\omega t-kx)$$. The graph in the figure shows how the displacement $$y$$ at a fixed point varies with time $$t$$. Which one of the labeled points shows a displacement equal to that at the position $$x=\pi/2k$$ at time $$t=0$$

A
$$P$$

B
$$Q$$

C
$$R$$

D
$$S$$

Solution

Given ,
$$Y=a_{0}\sin (\omega t-kx)$$.
Now, at $$x=\dfrac{\pi}{2k} \ and\ t=0$$
we get,
$$Y=a_{0}\sin (\omega (0)-k(\dfrac{\pi }{2k})$$.

$$Y=-a_0$$
from the figure point (Q) shows a displacement equal to (-a_0).
hence, option B is correct
Question 6
1 / -0

A simple harmonic plane wave propagates along x-axis in a medium. The displacement of the particles as a function of time is shown in figure, for $$x = 0$$ (curve 1) and $$x = 7$$ (curve 2).
The two particles are within a span of one wavelength.
The speed of the wave is

A
$$12 \,m/s$$

B
$$24 \,m/s$$

C
$$8 \,m/s$$

D
$$16 \,m/s$$

Solution

Let general wave equation is $$y = A \,sin (\omega t - kx + \phi)$$
$$v = \dfrac {dy}{dt} = A \omega cos (\omega t - kx + \phi)$$
for curve $$(1 ) , x = 0$$
at $$t = 0, x = 0 ,$$ we have $$ y = 0$$
$$\Rightarrow 0 = A \,sin[\phi] \Rightarrow sin \phi = 0$$
$$\Rightarrow \phi = 0 \,or \,\pi$$
here $$\phi = \pi$$ (because velocity is negative)
for curve $$(2), x = 7 \,cm$$
at $$t = 0, x = 7 \,cm, y = -1$$
$$-1 = sin(-k \times 7 + \pi)$$
$$\Rightarrow sin(-7k + \pi) = -1/2$$
$$\Rightarrow -7k + \pi = 2n\pi + \dfrac {7\pi}{6}$$ or $$2n\pi + \dfrac {11\pi}{6}$$
here $$\Rightarrow -7k + \pi = 2n\pi + \dfrac {11\pi}{6}$$
(because at t = 0, velocity is positive)
$$\Rightarrow -7 \left ( \dfrac {2\pi}{\lambda} \right ) = 2n\pi + \dfrac {5\pi}{6}$$
$$\Rightarrow \lambda = \dfrac {-14\pi}{5\pi/6 + 2n\pi}$$
$$\Rightarrow \lambda = \dfrac {-84}{12n + 5}$$
for $$n = -1, \lambda = 12 \,cm$$
for $$n = -2, \lambda = \dfrac {84}{19}cm$$ (not possible)
because $$\lambda > 7 \,cm$$
$$v = f\lambda = 100 \times \dfrac {12}{100} = 12 \,m/s$$
Question 7
1 / -0

Two separated sources emit sinusoidal travelling waves but have the same wavelength $$ \lambda $$ and are in phase at their respective sources. One travels a distance $$ l_{1} $$ to get to the observation point while the other travels a distance $$ l_{2} $$. The amplitude is minimum at the observation point, if $$ l_{1}-l_{2} $$ is an

A
odd integral multiple of $$ \lambda $$

B
even integral multiple of $$ \lambda $$

C
odd integral multiple of $$ \lambda / 2 $$

D
odd integral multiple of $$ \lambda / 4 $$

Solution
Question 8
1 / -0

Two vibration strings of the same material but lengths L and 2L have radii 2r and r, respectively. The are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length L with frequency $$n_1$$ and the other with frequency $$n_2$$. The ratio $$n_1/n_2$$ is given by

A
2

B
4

C
8

D
1

Solution

$$n_{1}=\frac{1}{2l}\sqrt{\left [ \frac{T}{4\pi r^{2}\rho } \right ]}$$

and $$n_{2}=\frac{1}{4l}\sqrt{\left [ \frac{T}{\pi r^{2}\rho } \right ]}$$

$$\therefore \frac{n_{1}}{n_{2}}=2\times \frac{1}{2}=1$$
Question 9
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Figure 5.55 shows a student setting up wave on a long stretched string. The student's hand makes one complete up and down movement in $$0.4 \,s$$ and in each up and down movement the hand moves by a height of $$0.3 \,m.$$ The wavelength of the waves on the string is $$0.8 \,m.$$
The amplitude of the wave is

A
$$0.15 \,m$$

B
$$0.3 \,m$$

C
$$0.075 \,m$$

D
cannot be predicted

Solution

$$Frequency = \dfrac {1}{Time \,Period} = \dfrac {1}{0.4} = 2.5 \,Hz$$
$$Amplitude = \dfrac {1}{2} \times 0.3 \,m = 0.15 \,m$$
Question 10
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Four pieces of string each of length L are joined end to end to make a long string of length $$4L.$$ The linear mass density of the strings are $$\mu, 4\mu, 9\mu$$ and $$16\mu,$$ respectively. One end of the combined string is tied to a fixed support and a transverse wave has been generated at the other end having frequency $$f$$ (ignore any reflection and absorptions). String has been stretched under a tension $$F.$$
Find the ratio of wavelengths of the waves on four strings, starting from right hand side.

A
$$12 : 6 : 4 : 3$$

B
$$4 : 3 : 2 : 1$$

C
$$3 : 4 : 6 : 12$$

D
$$1 : 2 : 3 : 4$$

Solution

Frequency of wave is same on all the four strings. So,
$$\lambda_1 = \dfrac {v_1}{f},$$ $$\lambda_2 = \dfrac {v_2}{f},$$
$$\lambda_3 = \dfrac {v_3}{f},$$ $$\lambda_4 = \dfrac {v_4}{f},$$
$$\lambda_4 : \lambda_3 : \lambda_2 : \lambda_1 = \dfrac {1}{4} : \dfrac {1}{3} : \dfrac {1}{2} : 1 = 3 : 4 : 6 : 12$$