Waves Test

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Waves Test - 54

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$1 : 1$$

B
$$2 : 1$$

C
$$2 : \sqrt {3}$$

D
$$\sqrt {3} : 2$$

Solution
Question 2
1 / -0

A particle performing $$S.H.M$$ is found at its equilibrium at $$t= 1\ s$$ and it is found to have a speed of $$0.25\ m/s$$ at $$t=2\ s$$.If the period of oscillation is $$6\ s$$.Calculate amplitude of oscillation

A
$$\dfrac { 3 }{ 2\pi } m$$

B
$$\dfrac { 3 }{ 4\pi } m$$

C
$$\dfrac { 6 }{ \pi } m$$

D
$$\dfrac { 3 }{ 8\pi } m$$

Solution

Given,
$$\omega=\dfrac{2\pi}{6}=\dfrac{\pi}{3}$$
Let the displacement of particle be,

$$y=A\sin(\omega t+\phi)$$

At $$ t=1,y=0$$

$$\dfrac{\pi t}{3}+\phi=\pi$$

$$\phi=\dfrac{2\pi}{3}$$

Velocity at $$t=1, 0.25m/s$$

$$v=A\omega\cos\left(\dfrac{2\pi}{3}+\dfrac{2\pi}{3}\right)=\dfrac{A\omega}{2}$$

$$0.25=\dfrac{\pi A}{6}$$

$$A=\dfrac{3}{2\pi}$$

Option $$\textbf A$$ is the correct answer
Question 3
1 / -0

The equation of a progressive wave are
$$Y=sin[200n(t-\cfrac x {330})]$$,where x is in meter and f is in second. The frequency and velocity of wave are

A
$$100 Hz,5m/s$$

B
$$300 Hz,100m/s$$

C
$$100 Hz,330 m/s$$

D
$$30 m/s,5Hz$$

Solution
Question 4
1 / -0

A $$4$$ kg particle is moving along the x-axis under the action of the force $$F=-\left(\dfrac{\pi^2}{16}\right)\times N$$. At $$t=2$$sec, the particle passes through the origin and at $$t=10$$ sec its speed is $$4\sqrt{2}$$m/s. The amplitude of the motion is?

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

B
$$\dfrac{16}{\pi}$$m

C
$$\dfrac{4}{\pi}$$m

D
$$\dfrac{16\sqrt{2}}{\pi}$$m

Solution
Question 5
1 / -0

Three waves of amplitude $$10\mu\ m,4\mu\ m$$ and $$7\mu\ m$$ arrive at a point with successive phase difference of $$\pi/2$$. The amplitude of the resultant wave is

A
$$2\mu m$$

B
$$7\mu m$$

C
$$5\mu m$$

D
$$1$$

Solution

Given,

Amplitude of the waves are,

$$a_1=10\mu m,\,a_2=4\mu m,\,a_3=7\mu m$$

and the phase difference between 1st and 2nd wave is $$\dfrac{\pi}{2}$$ and that between 2nd and 3rd wave is $$\dfrac{\pi}{2}$$. Therefore, phase difference between 1st and 3rd is $$\pi$$.

Combining 1st with 3rd, their resultant amplitude will be

$$A_1^2=a_1^2+a_3^2+2a_1a_3cos\phi$$

or

$$A_1=\sqrt{10^2+7^2+2\times 10\times 7cos\pi}$$

$$=\sqrt{100+49-140}=\sqrt 9 =3\mu m$$ in the direction of first.

Now combining this with 2nd wave we get the resultant wave,

$$A^2=A_1^2+a_2^2+2A_1a_2 \dfrac{cos\pi}{2}$$

or

$$A=\sqrt{3^2+4^2+2\times 3\times 4cos90^2}=\sqrt{9+16}=5\,\mu m$$
Question 6
1 / -0

An object of mass m is attached to a spring.The restoring force of the spring is F=$$ - \lambda {x^3},$$ where x is the displacement. the oscillation period depends on the mass, $$\lambda $$ and oscillation amplitude.suppose the object is initially at rest.If the initial displacement is D then its period is $$\tau $$ .If the initial displacement is 2D, find the period.(Hint: Use dimension analysis.)

A
8$$\tau $$

B
2$$\tau $$

C
$$\tau $$

D
$$\tau $$/2
Question 7
1 / -0

A transverse progressive wave on a stretched string has a velocity of $$10ms^{-1}$$ and frequency of $$100Hz$$. The phase difference between two particles of the string which nbare $$2.5cm$$ apart will be :

A
$$\cfrac{\pi}{8}$$

B
$$\cfrac{\pi}{4}$$

C
$$\cfrac{3\pi}{8}$$

D
$$\cfrac{\pi}{2}$$

Solution

Given,

$$v=10m/s$$ velocity

$$f=100Hz$$ frequency

$$\Delta=2.5cm$$ path difference

wavelength, $$\lambda=\dfrac{v}{f}$$

$$\lambda=\dfrac{10}{100}=0.1m$$

Phase difference,

$$\phi=\dfrac{2\pi}{\lambda}\times \Delta$$

$$\phi=\dfrac{2\pi}{0.1}\times 2.5\times 10^{-2}$$

$$\phi=\dfrac{\pi}{2}$$

The correct option is D.
Question 8
1 / -0

Equation of two $$S.H.M,\,\,{x_1} = 5\sin \left( {2\pi t + \pi /4} \right),\,{x_2} = 5$$ $$\sqrt 2 \left( {\sin 2\pi t + \cos 2\pi t} \right)$$. Ratio of amplitude & phase difference will be :

A
$$2:\,1,\,0$$

B
$$1:\,2,\,0$$

C
$$1:2,\,\pi /2$$

D
$$2:1,\,\pi /2$$

Solution

Equation of S.H.M ,

$$x_1=5sin(2\pi t +\pi/4)$$. . . . . . .(1)

$$x_2=5\sqrt{2}(sin2\pi t+cos2\pi t)$$

$$x_2=10(\dfrac{1}{\sqrt{2}}.sin2\pi t+\dfrac{1}{\sqrt{2}}.cos2\pi t)$$

$$x_2=10(cos\dfrac{\pi}{4}.sin2\pi t+sin\dfrac{\pi}{4}.cos2\pi t)$$

$$x_2=10(sin2\pi t+\pi/4)$$. . . . . . . . ..(2)

The ratio of amplitude, $$a_1:a_2=1:2$$

The phase difference is, $$\phi=\theta _1-\theta_2$$

$$\phi=\dfrac{\pi}{4}-\dfrac{\pi}{4}=0$$

The correct option is B.
Question 9
1 / -0

A wave is travelling along a string. At an instant shape of the string is as shown in figure. At this instant, point A is moving upwards. Which of the following statements is/are correct ?

A
The wave is travelling to the right

B
Displacement amplitude of the wave is equal to displacement of B at this instant

C
At this instant velocity of C is also directed upwards

D
Phase difference between A and C may be equal to $$\pi/2$$

Solution

$$\textbf{Explanation}$$
1) The velocity of $$A$$ is upward only if Wave moves to the left

2) Displacement of $$B$$ from mean position is the Amplitude of wave.

3) as Wave is propogating towards left, $$v_{c}$$ is downwards.

4) We don't know distance between $$A, C$$
as $$\triangle \phi = \dfrac {2\pi}{\lambda} (\triangle x)$$
$$\therefore \triangle \phi$$ cannot be known
Answer B.
Question 10
1 / -0

In a string the speed of wave is $$10m/s$$ and its frequency is $$100$$ Hz. The value of the phase difference at a distance $$2.5$$cm will be :

A
$$\pi/2$$

B
$$\pi/8$$

C
$$3\pi/2$$

D
$$4\pi$$

Solution

Given,

$$f=100Hz$$

$$x=2.5cm$$

$$v=10m/s$$ path difference

wavelength, $$\lambda =\dfrac{v}{f}=\dfrac{10}{100}=0.1m$$

The phase difference, $$\phi=\dfrac{2\pi}{\lambda}\times x$$

$$\phi=\dfrac{2\pi}{0.1}\times 2.5\times 10^{-2}$$

$$\phi=\dfrac{\pi}{2}$$

The correct option is A.