Waves Test

Result

Waves Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

A particle executes SHM of amplitude A if $$T_1$$ and $$T_2$$ are the time taken by the particle to traverse from 0 to A/2 and from A/2 to A respectively. Then $$T_1/T_2$$ will be equal to

A
1

B
1/2

C
1/4

D
2

Solution
Question 2
1 / -0

For a wave propagating in a medium, identify the property that is independent of the others

A
Velocity

B
Wavelength

C
Frequency

D
All these depend on each other

Solution

Any medium has its characteristic such as refractive index which affects the characteristics of a wave such as wavelength, velocity etc. ... The velocity and wavelength of a wave decrease by a factor p, (refractive index) on passing through a medium but frequency remains unchanged.
Question 3
1 / -0

A uniform rope of length $$L$$ and mass $${m_1}$$ hangs vertically from a rigid support . A block of mass $${m_2}$$ is attached to the free end of the rope. A transverse pulse of wavelength $${\lambda _1}$$ is produced at the lower end of the rope . the wavelength of the pulse when it reaches the top of the rope is $${\lambda _2}$$. The ratio $$\frac{{{\lambda _1}}}{{{\lambda _2}}}$$ is:

A
$$\sqrt {\dfrac{{{m_1}}}{{{m_2}}}} $$

B
$$\sqrt {\dfrac{{{m_1} + {m_2}}}{{{m_2}}}} $$

C
$$\sqrt {\dfrac{{{m_2}}}{{{m_1}}}} $$

D
$$\sqrt {\dfrac{{{m_1} + {m_2}}}{{{m_1}}}} $$

Solution
Question 4
1 / -0

The amplitude of a $$SHM$$ reduces to $$1/3$$ in first $$20$$ second then in first $$40$$ second its amplitude becomes:

A
$$\dfrac{1}{3}$$

B
$$\dfrac{1}{9}$$

C
$$\dfrac{1}{27}$$

D
$$\dfrac{1}{{\sqrt 3 }}$$

Solution
Question 5
1 / -0

A particle of mass $$m$$ oscillates along $$x-$$axis according to equations $$x=a\sin \omega t$$. The nature of the graph between momentum and displacement of the particle is

A
Straight line passing through origin

B
Circle

C
Hyperbola

D
Ellipse

Solution
Question 6
1 / -0

The graph shown the variation of displacement of a particle executing SHM with time. We inference from this graph that-

A
the force is zero at time $$\dfrac{3T}{4}$$

B
the velocity is maximum at time $$\dfrac{T}{2}$$

C
the acceleration is maximum at time $$T$$

D
the P.E. is equal to half of total energy at time $$\dfrac{T}{2}$$

Solution

As the graph shows the dissplacement vs time for a particle performing SHM,

At time $$\dfrac T2$$, the displacem,ent of particle is zero. So, it will be at the mean position of its SHM.

At time $$\dfrac{T}{2};v=max$$
$$\therefore $$ Total energy will be stored in form of the kinetic energy.

So, at $$\dfrac T2$$, the velocity will be maximum.
Question 7
1 / -0

Two Waves of amplitudes $${ A }_{ 0 }$$ and $$x{ A }_{ 0 } $$ pass through a region. If x >1, the difference in the maximum and minimum resultant amplitude possible is

A
$$(x+1){ A }_{ 0 }$$

B
$$(x-1){ A }_{ 0 }$$

C
$$2x{ A }_{ 0 }$$

D
$$2{ A }_{ 0 }$$

Solution
Question 8
1 / -0

A metal rod 40 cm long is dropped onto a wooden floor and rebounds into air. Compressional waves of many frequencies are thereby set up in the rod. If the speed of compressional waves in the rod is 5500 $$m s ^ { - 1 }$$ what is the - lowest frequency of compressional waves to which the rod resonates as it rebounds?

A
6875 Hz

B
16875 Hz

C
675 Hz

D
0.11 Hz

Solution

$$L = 2\left( {\frac{\lambda }{4}} \right)$$
or$$,$$ $$\lambda = 2L = 2\left( {0.40m} \right) = 0.80m$$
then$$,$$ from $$\lambda = VT = \frac{V}{F}$$
$$F = \dfrac{V}{\lambda } = \dfrac{{5500}}{{0.80}} = 6875\,Hz$$
Hence,
option $$(A)$$ is correct answer.
Question 9
1 / -0

Consider the wave represented by $$y=\cos(500t-70x)$$ where $$x$$ is in metres and $$t$$ in seconds. the two nearest points in the same phase have a separation of

A
$$2\pi/7\ m$$

B
$$2\pi/7\ cm$$

C
$$20\pi/7\ m$$

D
$$20\pi/7\ cm$$

Solution
Question 10
1 / -0

A particle is executing simple harmonic motion between extreme positions given by (-1, -2, -3)cm and (1, 2, 1)cm. Its amplitude of oscillation is

A
6 cm

B
4 cm

C
2 cm

D
3 cm

Solution