Waves Test

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Waves Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

A piece of cork is floating on water in a small tank. The cork oscillates up and down vertically when small ripples pass over the surface of water. The velocity of the ripples being $$0.21\ ms^{-1}$$, wave length 15 mm and amplitude 5 mm, the maximum velocity of the piece of cork is $$\displaystyle\bigg (\pi = \frac{22}{7}\bigg )$$

A
$$0.44 \ ms^{-1}$$

B
$$0.24 \ ms^{-1}$$

C
$$2.4 \ ms^{-1}$$

D
$$4.4 \ ms^{-1}$$

Solution

We know that in wave motion , maximum particle velocity is given by ,
$$v_{max}=\omega A=2\pi f A$$ ,
where $$\omega=$$ angular frequency ,
$$f=$$ linear frequency ,
$$A=$$ amplitude ,
given - wave velocity ,$$u=0.21m/s , \lambda=15mm=0.015m , A=5mm=0.005m$$ ,
by , $$f=u/\lambda=0.21/0.015$$ Hz ,
therefore maximum particle (cork) velocity ,
$$v_{max}=2\times(22/7)\times(0.21/0.015)\times0.005=0.44m/s$$
Question 2
1 / -0

Three waves of equal frequency having amplitudes $$10\ \displaystyle \mu m$$, $$ 4\ \displaystyle \mu m$$ and $$7\ \displaystyle \mu m$$ arrive at a given point with a successive phase difference of $$\displaystyle \pi /2$$. The amplitude of the resulting wave is $$\displaystyle \mu m$$ in given by

A
$$7$$

B
$$6$$

C
$$5$$

D
$$4$$

Solution

The wave $$1$$ & $$3$$ out of phase. Hence resultant phase difference between then is $$\pi $$
$$\therefore $$Resultant amplitude of $$1$$ & $$3$$=$$10-7$$=$$3\mu m$$
This wave has phase difference of $$\frac { \pi }{ 2 } $$ with $$4\mu m$$
$$\therefore $$Resultant amplitude
$$=\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } \\ =5\mu m$$
Hence option (C) is correct
Question 3
1 / -0

A wave of wavelength 4 mm is produced in air and it travels at a speed of 300 m/s. Will it be audible?

A
$$20,000Hz$$

B
$$70000Hz$$

C
$$92000Hz$$

D
$$90000HZ$$

Solution

We have , $$v=f\lambda$$ ,
or $$f=v/\lambda$$ ,
given $$\lambda=4mm=0.004m , v=300m/s$$ ,
therefore , frequency
$$f=300/0.004=75000Hz$$ ,
this frequency is not audible because audible frequency range is from 20Hz to 20,000Hz , and 75000Hz is above 20,000Hz.
Question 4
1 / -0

Two waves of amplitude $$\displaystyle A_{1}\: $$and$$\: A_{2} $$ respectively and equal frequency travels towards the same point. The amplitude of the resultant wave is

A
$$\displaystyle A_{1}+A_{2} $$

B
$$\displaystyle A_{1}-A_{2} $$

C
between $$\displaystyle A_{1}+A_{2} $$ and $$\displaystyle A_{1}-A_{2} $$

D
Can not say

Solution

If the two waves are in constructive phase, then the resultant amplitude will be equal to
$$A={ A }_{ 1 }+{ A }_{ 2 }$$ (maximum)
If they are in destructive phase, then
$$A={ A }_{ 1 }-{ A }_{ 2 }$$
Therefore the resultant amplitude will be always between $${ A }_{ 1 }+{ A }_{ 2 }$$ & $${ A }_{ 1 }-{ A }_{ 2 }$$
Hence option (C) is correct.
The resultant will depend on their phase difference.
Question 5
1 / -0

A particle performing SHM on the y axis according to equation $$\displaystyle y=A+B\sin\omega t$$, its amplitude is

A
A

B
B

C
A + B

D
$$\displaystyle \sqrt{A^{2}+B^{2}}$$

Solution

$$y=A+Bsin\omega t$$
Here the mean position is A and the amplitude is $$B$$
In a single SHM where there is no superposition the term the coefficient of $$sin\omega t$$ is generally the amplitude.
Question 6
1 / -0

When a sound wave is reflected from a rigid wall, the phase difference between the reflected and incident wave

A
0

B
$$\pi$$

C
$$\pi /2$$

D
$$\pi /4$$

Solution

when a sound wave gets reflected from a rigid wall then it undergoes a phase change of $$\pi$$ w.r.t. to incident wave.
so the answer is B.
Question 7
1 / -0

A wave represented by the equation $$\displaystyle y=a\cos \left ( kx-\omega t \right )$$ is superimposed with another wave to form a stationary wave such that the point x = 0 is a node. The equation for other wave is

A
$$\displaystyle a\sin \left ( kx+\omega t \right )$$

B
$$\displaystyle -a\cos \left ( kx+\omega t \right )$$

C
$$\displaystyle -a\cos \left ( kx-\omega t \right )$$

D
$$\displaystyle -a\sin \left ( kx-\omega t \right )$$

Solution

To form a stationary wave with $$y=a\cos { \left( kx-wt \right) } $$ the wave should be in the opposite direction of this wave and amplitude should be negative that is the phase difference should be of $$\pi $$
$$y=a\cos { \left( kx+wt+\pi \right) } \\ \left[ y=-a\cos { \left( kx+wt \right) } \right] $$
Hence option (B) is correct
Question 8
1 / -0

Two sinusoidal waves of the same frequency travel in the same direction along a string. If $$\displaystyle A_{1}=3.0\: cm,A_{2}=4.0\: cm,\phi _{1}=0,$$ and $$\displaystyle \phi _{2}=\pi /2$$ rad, what is the amplitude of the resultant wave?

A
5 cm

B
6 cm

C
7 cm

D
8 cm

Solution

$${ A }_{ 1 }=3cm\\ { A }_{ 2 }=4cm\\ \Delta \phi =\frac { \pi }{ 2 } $$
Resultant amplitude will be
$$A=\sqrt { { A }_{ 1 }^{ 2 }+{ A }_{ 2 }^{ 2 }+2{ A }_{ 1 }{ A }_{ 2 }\cos { \left( \frac { \pi }{ 2 } \right) } } \\ A=\sqrt { { A }_{ 1 }^{ 2 }+{ A }_{ 2 }^{ 2 } } \\ A=\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } \\ \left[ A=5cm \right] $$
Hence option (A) is correct.
Question 9
1 / -0

Two waves passing through a region are represented by
$$\displaystyle y_{1}=5\: mm\sin [(2\pi \: cm^{-1})x-(50\pi s^{-1})t]$$
and $$\displaystyle y_{2}=10\: mm\sin [(\pi \: cm^{-1})x-(100\pi s^{-1})t]$$
Find the vertical displacement of a particle at x = 1 cm at time t = 5.0 ms

A
5 mm

B
6 mm

C
7 mm

D
8 mm

Solution

According to the principle of superposition each wave produces its disturbance independent of the other and he resultant disturbance is equal to the vector sum of the individual disturbance The displacements of the particle at x = 1 cm at time t = 5.0 ms due to the two waves are
$$\displaystyle y_{1}=5\: mm\sin [(2\pi \: cm^{-1})x-(50\pi s^{-1})t]$$
$$\displaystyle y_{1}=5\: mm\sin [(2\pi \: cm^{-1})\times 1cm-(50\pi s^{-1})5\times 10^{-3}sec]$$
$$\displaystyle =5\: \: mm\sin \left [ 2\pi -\frac{\pi }{4} \right ]=-5\: mm$$
and $$\displaystyle y_{2}=10mm\sin [(\pi cm^{-1})x-(100\pi s^{-1})t]$$
$$\displaystyle y_{2}=10mm\sin [(\pi cm^{-1})\times 1cm-(100\pi s^{-1})5\times 10^{-3}sec]$$
$$\displaystyle =10\: \: mm\: \sin \left [ \pi -\frac{\pi }{2} \right ]=10\: mm$$
The net displacement is $$\displaystyle y=y_{1}+y_{2}=10\: mm-5\: mm=5\: mm$$
Question 10
1 / -0

A ripple is created in water. The amplitude at a distance of $$5$$cm from the point where the sound ripple was created is $$4$$cm. Ignoring damping, what will be the amplitude at the distance of $$10$$cm.

A
$$\sqrt{16}$$cm

B
$$\sqrt{8}$$cm

C
$$\sqrt{4}$$cm

D
$$\sqrt{2}$$cm

Solution

Let's assume that the ripple was created in a pool of water at the center so that with time the ripple moves away.
Let the energy of the ripple be $$E$$ .
At a distance of $$5\ cm$$ from the point of origin the amplitude is $$4\ cm$$.
Energy distributed in circle of circumference $$2\pi 5 \ cm$$.
Energy per unit length, $$E_1=\dfrac{E}{10\pi}$$
When it reaches $$10\ cm$$, $$E_2=\dfrac{E}{2\pi \times 10}$$ $$=\dfrac{E_1}{2}$$
Now energy is proportional to square of amplitude ($$A$$).
Thus if per unit energy decreases by $$\dfrac{1}{2}$$, then amplitude will decrease by $$\dfrac{1}{\sqrt{2}}$$.
New amplitude, $$A_1=\dfrac{4}{\sqrt{2}}=\sqrt{8}$$

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Waves Test - 33

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Waves Test - 33

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SHARING IS CARING

Question 1

$$0.44 \ ms^{-1}$$

$$0.24 \ ms^{-1}$$

$$2.4 \ ms^{-1}$$

$$4.4 \ ms^{-1}$$

Solution

Question 2

$$7$$

$$6$$

$$5$$

$$4$$

Solution

Question 3

$$20,000Hz$$

$$70000Hz$$

$$92000Hz$$

$$90000HZ$$

Solution

Question 4

$$\displaystyle A_{1}+A_{2} $$

$$\displaystyle A_{1}-A_{2} $$

between $$\displaystyle A_{1}+A_{2} $$ and $$\displaystyle A_{1}-A_{2} $$

Can not say

Solution

Question 5

A

B

A + B

$$\displaystyle \sqrt{A^{2}+B^{2}}$$

Solution

Question 6

0

$$\pi$$

$$\pi /2$$

$$\pi /4$$

Solution

Question 7

$$\displaystyle a\sin \left ( kx+\omega t \right )$$

$$\displaystyle -a\cos \left ( kx+\omega t \right )$$

$$\displaystyle -a\cos \left ( kx-\omega t \right )$$

$$\displaystyle -a\sin \left ( kx-\omega t \right )$$

Solution

Question 8

5 cm

6 cm

7 cm

8 cm

Solution

Question 9

5 mm

6 mm

7 mm

8 mm

Solution

Question 10

$$\sqrt{16}$$cm

$$\sqrt{8}$$cm

$$\sqrt{4}$$cm

$$\sqrt{2}$$cm

Solution

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