Waves Test

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Waves Test - 73

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Weekly Quiz Competition

Question 1
1 / -0

A pulse is started at a time t = 0 along the +x direction on a long, taut string. The shape of the pulse at t = 0 is given by function y with
$$y=\begin{cases} \dfrac { x }{ 4 } +1\quad \ for\quad -4 < x \le 0 \\ -x+1\quad for\quad 0 < x < 1 \\ 0 in any other case\end{cases}$$
here y and x are centimeters. The linear mass density of the string is 50 g/m and it is under a tension of 5N.
The shape of the string is drawn at t = 0 and the area of the pulse enclosed by the string and the x-axis is measured. It will be equal to

A
$$2\ {cm}^{2}$$

B
$$2.5\ {cm}^{2}$$

C
$$4\ {cm}^{2}$$

D
$$5\ {cm}^{2}$$

Solution
Question 2
1 / -0

The phase difference between two waves, represented by $${ y }_{ 1 }={ 10 }^{ -6 }\sin { \left[ 100t+\left( x/50 \right) +0.5 \right]\ m } $$ and $${ y }_{ 2 }={ 10 }^{ -6 }\cos { \left[ 100t+\left( x/50 \right) \right]\ m } $$. Where $$x$$ is expressed in metre and $$t$$ is expressed in seconds, is approximately

A
$$1.07\ radian$$

B
$$2.07\ radian$$

C
$$0.5\ radian$$

D
$$1.5\ radian$$

Solution
Question 3
1 / -0

The amplitude of a damped harmonic oscillator becomes halved in $$1$$ minute. After three minutes the amplitude will become $$1/x$$ of initial amplitude where $$x$$ is:

A
$$2\times 3$$

B
$${2}^{2}$$

C
$${2}^{3}$$

D
$$3\times {2}^{2}$$

Solution
Question 4
1 / -0

Displacement of particles in a string in x-direction and is represented by y. Account the following expression for y, those describing wave motion are.

A
$$\cos kx\sin\omega t$$

B
$$k^2x^2-\omega^2 t^2$$

C
$$\cos^2(kx+\omega t)$$

D
$$\cos (kx^2-\omega^2 t^2)$$
Question 5
1 / -0

A string fixed at one end only is vibrating in its third harmonic. The wave function is $$y(x,t) = 0.02 sin(3.13x) cos(512t)$$, where y and x are in metres and t is in seconds. The nodes are formed at positions

A
(0 m, 2 m)

B
(0.5 m, 1.5 m)

C
(0 m, 1.5 m)

D
(0.5 m, 2 m)

Solution

Node in formed only at the finest end of the string and the free end acts as an anti node.

In the third harmonics two nodes are formed:

$$y\left( x,t \right) =0.02\sin { \left( 3.13x \right) \cos { \left( 512t \right) } } $$

Standard equation given by

$$y\left( x,t \right) =2a\sin { \left( \cfrac { 2\pi }{ \lambda } x \right) } \cos { \left( 2\pi vt \right) } $$

Comparing both equation, we get

$$3.13x\quad =\cfrac { 2\pi }{ \lambda } x\\ or,\quad \lambda =\cfrac { 2\lambda }{ 3.13 } \\ \quad \quad \quad \quad =2 m (approx)$$

The nodes are formed at $$\cfrac { \lambda }{ 4 } =0.5$$ from origin and at $$\cfrac { 3\lambda }{ 4 } =1.5$$ from origin.
Question 6
1 / -0

A wave $$10\sin{(ax+bt)}$$ is reflected from dense medium at an origin. If 81% of energy is reflected then the equation of reflected wave is

A
$$y=-8.1\sin{(ax-bt)}$$

B
$$y=-8.1\sin{(ax+bt)}$$

C
$$y=-9\sin{(ax-bt)}$$

D
$$y=-10\sin{(ax-bt)}$$

Solution

Wave equ before reaction : $$10sin(ax+bt)$$
Intensity of the reflected wave $$=81$$%
$$=0.81$$ of the original.
Since $$I\alpha { A }^{ 2 }$$
Amplitude of reflected wave $$=\sqrt { I } =\sqrt { 0.81 } =0.9$$ of the original $$A$$.
$$\therefore \quad { A }^{ 1 }=0.9A$$
$${ A }^{ 1 }=0.9\times 10=9$$
Equ of reflected wave :
$$\left[ y=-9\sin\left( ax-bt \right) \right] $$

$$\therefore $$ Option (C) is correct.
Question 7
1 / -0

A tray of mass M = $$10$$kg is supported on two identical springs, each of spring constant k, as shown in figure. When the tray is depressed a little and released, it executes simple harmonic motion of period $$1.5$$s. When a block of mass m is placed on the tray, the period of oscillation becomes $$3$$s.The value of m is

A
$$10$$kg

B
$$20$$kg

C
$$30$$kg

D
$$40$$kg

Solution

$$T_1=2\pi \sqrt{\cfrac{M}{2K}}$$
$$T_2=2\pi \sqrt{\cfrac{M+m}{2K}}$$
$$\Longrightarrow \cfrac{T_1}{T_2}=\cfrac{\sqrt{M}}{\sqrt{M+m}}$$
$$\Longrightarrow \cfrac{1.5}{3}=\sqrt{\cfrac{10}{10+m}}$$
$$\Longrightarrow \cfrac{1}{4}=\cfrac{10}{m+10}$$
$$\Longrightarrow m=30kg.$$
Question 8
1 / -0

To determine the position of a point like object precisely ______ light should be used.

A
polarized

B
short wavelength

C
long wavelength

D
intense

Solution

To determine the position of a point like object precisely light of short wavelength should be used.
Question 9
1 / -0

A wave frequency $$100Hz$$ travels along a string towards its fixed end. When this wave travels back after reflection, a node is formed at a distance of $$10cm$$ from the fixed end. The speed of the wave (incident and reflected) is

A
$$5m/s$$

B
$$10m/s$$

C
$$20m/s$$

D
$$40m/s$$

Solution

Frequency of the wave $$=100{ H }_{ 3 }$$
$$\therefore $$ Time period $$=\dfrac { 1 }{ f } =\dfrac { 1 }{ 100 } =0.01sec$$
wavelength of the wave $$=2\times 10cm=20cm$$
$$\lambda =0.2m$$
Speed of the wave $${ V }_{ w }=\dfrac { w }{ k } =\dfrac { 2\pi /T }{ 2\pi /\lambda } =\dfrac { \lambda }{ T } =\dfrac { 0.20 }{ 0.01 } \quad \left[ { V }_{ w }=20m/s \right] $$

$$\therefore $$ Option (C) is correct.
Question 10
1 / -0

The equation of a plane progressive wave is $$y = 0.02 \sin 8 \pi[t - \dfrac{x}{20}].$$ When it is reflected at a rarer medium, its amplitude becomes 75% of its previous value. The equation of the reflected wave is

A
$$y = 0.02 \sin 8 \pi [t - \dfrac{x}{20}]$$

B
$$y = 0.02 \sin 8 \pi [t + \dfrac{x}{20}]$$

C
$$y = 0.15 \sin 8 \pi [t + \dfrac{x}{20}]$$

D
$$y = 0.15 \sin 8 \pi [t - \dfrac{x}{20}]$$