Waves Test

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Waves Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

Two simple harmonic motions are represented by equations
$$ y _ { 1 } = 10 \sin ( 3 \pi t + \pi / 4 )$$ and
$$y _ { 2 } = 5 ( \sin 3 \pi t + \sqrt { 3 } \cos 3 \pi t ) \times 2 $$
Their amplitudes are in the ratio of

A
2

B
1

C
3

D
4

Solution
Question 2
1 / -0

The equation of a stationary wave is $$y=0.8cos\left( \dfrac { \pi x }{ 20 } \right) \sin 200\pi t$$ where x is in cm and t is in seconds. The separation between consecutive nodes is

A
10 cm

B
20 cm

C
30 cm

D
40 cm

Solution
Question 3
1 / -0

The displacement of a particle is given by x = 3 sin $$\left( 5\pi t \right) $$+ 4 cos$$\left( 5\pi t \right) $$ The amplitude of particle is

A
3

B
4

C
5

D
7

Solution
Question 4
1 / -0

The displacement of an oscillating particle varies with time (in seconds) according to the equations $$y(cm)=sin\dfrac { \pi }{ 2} (\dfrac { t }{ 2}+\dfrac { 1 }{ 3} )$$. The maximum acceleration of the particle is approximately.

A
$$5.21cm/s^2$$

B
$$3.62cm/s^2$$

C
$$1.81cm/s^2$$

D
$$0.62cm/s^2$$

Solution

From the given equation of the oscillation, we obtain,
The angular frequency of the oscillation $$\omega =\dfrac{\pi}{4}$$
The amplitude of the particle is $$a=1$$

So, the maximum acceleration of the particle under oscillation is given as:
$$a_{max}=\omega^2 a =\left(\dfrac{\pi}{4}\right)^2 \times1$$

$$=0.62\ cm/sec^2$$ [$$\because a=1$$]
Question 5
1 / -0

A tunnel is dug n the earth across one of its diameter. Two masses 'm' & '2m' are dropped from the ends of the tunnel. The masses collide and stick to each other and performs S.H.M. Then amplitude of S.H.M. will be : [R=radius of the earth]

A
R

B
R/2

C
$$\dfrac{{\sqrt {2R} }}{3}$$

D
2R/3

Solution

For mass $$'m'$$
$$\begin{array}{l} ui+ki=uf+kf \\ \Rightarrow \dfrac { { -GMm } }{ R } +0=\dfrac { { -GM } }{ { 2{ R^{ 3 } } } } \left( { 3{ R^{ 2 } }-0 } \right) m+\dfrac { 1 }{ 2 } m{ v^{ 2 } } \\ \Rightarrow \dfrac { { -GMm } }{ R } =\dfrac { { -3GM } }{ { 2R } } +\dfrac { { { V^{ 2 } } } }{ 2 } \\ \Rightarrow \dfrac { { -GM } }{ R } =\dfrac { { { V^{ 2 } } } }{ 2 } \\ \Rightarrow { V^{ 2 } }=\dfrac { { 2GM } }{ R } \\ \Rightarrow V=\sqrt { \dfrac { { 2GM } }{ R } } \end{array}$$
Conserving momentum before and after collinear.
Let $${V_{cm}}$$ is the velocity of combines mass.
$${V_{cm}} = \dfrac{{2mv - mv}}{{3m}} = \dfrac{v}{3}$$
Let the amplitude is $$x$$.
$$\begin{array}{l} ui+ki=uf+kf \\ \Rightarrow \dfrac { { -3GM\left( { 3m } \right) } }{ { 2R } } +\dfrac { 1 }{ 2 } \times 3m\times \dfrac { { { v^{ 2 } } } }{ 9 } =\dfrac { { -GM } }{ { 2{ R^{ 3 } } } } \left( { 3{ R^{ 2 } }-{ x^{ 2 } } } \right) \left( { 3m } \right) +0 \\ \Rightarrow x=\dfrac { { \sqrt { 2R } } }{ 3 } \end{array}$$
$$\therefore$$ AMplitude is $$\dfrac{{\sqrt {2R} }}{3}$$
$$\therefore$$ Option $$C$$ is correct answer.
Question 6
1 / -0

Equation of motion in the same direction is given by $$y_1 = A \ sin(\omega t - kx), y_2 = A \ sin(\omega t - kx - \theta)$$. The amplitude of the medium particle will be

A
2 A $$cos\frac{\theta}{2}$$

B
2 A $$cos \theta$$

C
A $$cos \theta$$

D
A $$cos\frac{\theta}{2}$$

Solution
Question 7
1 / -0

A body executing simple harmonic motion has a maximum acceleration equal to $$24metre/sec^2$$ and maximum velocity equal to $$16meter/sec$$. The amplitude of simple harmonic motion is:

A
$$\dfrac { 2 }{ 32} m$$

B
$$\dfrac { 32}{ 3}m $$

C
$$(\dfrac { 64 }{ 9 } )m$$

D
$$(\dfrac { 1024 }{ 9 } )m$$

Solution
Question 8
1 / -0

The amplitude of a particle executing $$SHM$$ is $$4cm$$. At the mean position the speed of the particle is $$16cm/sec$$. The distance of the particle from the mean position at which the speed of the particle becomes $$8\sqrt { 3} cm/s$$, will be:

A
$$2\sqrt { 3} cm/s$$

B
$$\sqrt { 3} cm/s$$

C
$$1cm$$

D
$$2cm$$

Solution

At mean position velocity is maximum
i.e., $$v_{max}=\omega a \Rightarrow \omega =\dfrac{v_{max}}{a}=\dfrac{16}{4}=4$$

The velocity can also be represented as:
$$\therefore v=\omega \sqrt{a^2-y^2}$$

$$\Rightarrow 8\sqrt 3=4\sqrt{4^2-y^2}$$

$$\Rightarrow 192=16(16-y^2)\Rightarrow 12=16-y^2$$

$$\Rightarrow y=2\ cm$$.
Question 9
1 / -0

A travelling wave $$Y = A$$ sin $$( k x - \omega t + \theta )$$ passes from a heavier string to lighter string. The reflected has amplitude $$0.5$$ $$A$$. The junction of the string is $$x = 0$$. The equation of the reflected wave is:

A
$$y ^ { \prime } = 0.5 A \sin ( k x + \omega t + \theta )$$

B
$$y ^ { \prime } = - 0.5 A \sin ( k x + \omega t + \theta )$$

C
$$y ^ { \prime } = - 0.5 A \sin ( \omega t - k x - \theta )$$

D
$$y ^ { \prime } = - 0.5 A \sin ( k x + \omega t - \theta )$$

Solution

$$\begin{array}{l} Y=A\sin \left( { Kx-\omega t+\theta } \right) \\ y'=0.5A\sin \left( { Kx+\omega t+\theta +\pi } \right) \\ =-0.5A\sin \left( { Kx+\omega t+\theta } \right) \\ Hence, \\ option\, \, B\, \, is\, \, correct\, \, answer. \end{array}$$
Question 10
1 / -0

The amplitude of simple harmonic motion represented by the displacement equation y(cm) = 4(sin 5$$\pi$$t + $$\sqrt2$$cos 5$$\pi$$t) is :

A
4 cm

B
4$$\sqrt2$$ cm

C
4$$\sqrt3$$ cm

D
4($$\sqrt2 + 1) cm$$

Solution

$$ y = u (sin5\pi t+\sqrt{2}cos5\pi t)$$
$$ =u sin 5 \pi zt+4\sqrt{2}cos5\pi t$$
$$ y = A sinwt + B coswt$$
amplitude $$ = \sqrt{A^{2}+B^{2}}$$
$$ = \sqrt{(4)^{2}+(4\sqrt{2})^{2}}$$
$$ = 4\sqrt{3}$$
Option (c)