Waves Test

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Waves Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

A string of length $$20\ cm$$ and linear mass density $$0.4\ g/cm$$ is fixed at both ends and is kept under a tension of $$16 \ N$$. A wave pulse is produced at $$t = 0$$ near an end as shown in figure which travels towards the other end. The string have the shape shown in the figure again in $$ 2 \times 10^{-x} sec$$. Find $$x$$.

A
$$1$$

B
$$2$$

C
$$8$$

D
$$3$$

Solution

Given : $$T =16$$ N $$L =0.2$$ m $$\mu =0.4$$ g/cm $$ = 0.04$$ kg/m
Velocity of wave $$v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{16}{0.04}} =20$$ m/s
Time after which the string has the same shape $$t = \dfrac{2L}{v}$$
$$\therefore$$ $$t = \dfrac{2(0.2)}{20}= 2\times 10^{-2}$$ s $$\implies x = 2$$
Question 2
1 / -0

Two waves are travelling in the same direction along a stretched string. The waves are $$90^0$$ out of phase. Each wave has an amplitude of 4.0 cm. The amplitude of the resultant wave is $$5.66 \times 10^{-x} m$$. Find x.

A
1

B
2

C
8

D
4

Solution

Let A be the amplitude of each wave. Thus the waves are
$$y_1=Acos(\omega t-kx)$$
and $$y_2=Asin(\omega t-kx)$$
Thus the net wave is $$y= y_1 + y_2 = Acos(\omega t-kx)+Asin(\omega t-kx)$$
$$=\sqrt{2}Asin(\omega t-kx+\dfrac{\pi}{4})$$
Thus the new amplitude is $$\sqrt{2}A=5.66cm = 5.66\times 10^{-2} m$$
Thus, x=2.
Question 3
1 / -0

Consider a sinusoidal travelling wave shown in figure. The wave velocity is $$+ 40 cm/s$$. Find the frequency.

A
$$2 Hz$$

B
$$6 Hz$$

C
$$8 Hz$$

D
$$10 Hz$$

Solution

Clearly from figure, the wavelength is equal to the distance between two in-phase points.
Let us consider the points where it cuts the x-axis.
The in-phase points cut x-axis at $$2.5\ cm$$ and $$6.5\ cm$$.
Thus the wavelength is $$4cm$$
We have $$v=\lambda\nu$$
$$\implies \nu=\dfrac{v}{\lambda}=\dfrac{40cm/s}{4cm}=10Hz$$
Question 4
1 / -0

The equation of a wave is given by $$Y\, =\, 5\, sin\, 10 \pi\, (t\, -\, 0.01x)$$ along the x-axis. (All the quantities are expressed in SI units}. The phase difference the points separated by a distance of 10 m along x-axis is

A
$$\displaystyle \frac{\pi}{2}$$

B
$$\pi$$

C
$$2 \pi$$

D
$$\displaystyle \frac{\pi}{4}$$

Solution

For the given wave, $$k=0.1\pi=\dfrac{2\pi}{\lambda}$$
$$\implies \lambda=20m$$
Thus phase difference between two points separated by 10m is $$\dfrac{2\pi}{\lambda}(x_2-x_1)$$
$$=\dfrac{2\pi}{\lambda}\times 10m=\pi$$
Hence correct answer is option B.
Question 5
1 / -0

A source oscillates with a frequency 25 Hz and the wave propagates with 300 m/s. Two points A and B are located at distances 10 m and 16 m away from the source. The phase difference between A and B is

A
$$\displaystyle \frac{\pi}{4}$$

B
$$\displaystyle \frac{\pi}{2}$$

C
$$\pi$$

D
$$2 \pi$$

Solution

Wavelength of the wave=$$\lambda=\dfrac{v}{\nu}=\dfrac{300}{25}=12m$$
Distance between the two points=$$16m-10m=6m=\dfrac{\lambda}{2}$$
$$=\dfrac{2\pi}{\lambda}\dfrac{\lambda}{2}=\pi$$
Question 6
1 / -0

Three one dimensional mechanical waves in an elastic medium is given as and
$$y_1\,=\,3A \,sin\,(\omega\,t\,-\,kx)$$
$$y_2\,=\,A \,sin\,(\omega\,t\,-\,kx\,+\,\pi)$$
$$y_3\,=\,2A \,sin\,(\omega\,t\,+\,kx)$$
are superimposed with each other. The maximum displacement amplitude of the medium particle would be

A
4A

B
3A

C
2A

D
A

Solution

It is very clear from the question that the waves 1 and 2 will always interfere destructively since, they are completely out of phase with respect to each other.

Therefore, the resultant wave will always have an amplitude of 3A - A = 2A

Now, this wave has to interact with the third wave, constructively to produce a wave that has the maximum amplitude.

This would only be possible if the resultant wave has an amplitude of 2A + 2A = 4A
Question 7
1 / -0

The period of a particle in SHM is $$8 s$$. At $$t=0$$ it is in its equilibrium position. Find the ratio of the distance traveled in the first $$2 s$$ and the next $$2 s$$.

A
$$3$$

B
$$5$$

C
$$2$$

D
$$1$$

Solution

Distance traveled in first 2 sec=OP=A and
distance traveled in next 2 sec=PO=A
Again the two distances are same.
Question 8
1 / -0

Three coherent waves having amplitudes 12 mm, 6 mm and 4 mm arrive at a given point with successive phase difference of $$\pi/2$$. Then the amplitude of the resultant wave is

A
7 mm

B
10 mm

C
5 mm

D
4.8 mm

Solution

Since, it is given that the waves arrive at the same point with a phase difference of $$ \pi/2 $$

Therefore, the waves having an amplitude of 12mm and 4 mm are completely out of phase, and then their resultant amplitude is 12 - 4 = 8mm

Now, the 8mm and 6mm waves are $$ 90^o $$ out of phase so, their resultant amplitude would be given by Pythagorus theorem

$$ A^2 = 6^2 + 8^2 $$

Therefore, A = 10 mm
Question 9
1 / -0

A wave travels on a light string. The equation of the waves is $$Y\, = \,A\, sin\,(kx\,-\,\omega\,t+\,30^{\circ})$$. It is reflected from a heavy string tied to end of the light string at x = 0 . If 64% of the incident energy is reflected then the equation of the reflected wave is

A
$$Y\, =\,0.8 \,A\, sin\,(kx\,-\,\omega\,t\,+\,30^{\circ}\,+\,180^{\circ})$$

B
$$Y\, =\,0.8 \,A\, sin\,(kx\,+\,\omega\,t\,+\,30^{\circ}\,+\,180^{\circ})$$

C
$$Y\, =\,0.8 \,A\, sin\,(kx\,-\,\omega\,t\,-\,30^{\circ})$$

D
$$Y\, =\,0.8 \,A\, sin\,(kx\,-\,\omega\,t\,+\,30^{\circ})$$
Solution
There are three things we need to take into account:
- Energy transfer
- Change in velocity
- Change in phase
We know that power delivered is proportional to $$A^{2}$$
Hence if power(energy) reduces to 64%. We get that Amplitude must reduce to 80% or 0.8A.
Now the reflected wave is moving in the opposite direction. (velocity is negative now).
Also because of the hard soft boundary reflection (there is a phase lag of $$180^{\circ}$$
Hence the new equation becomes:
$$y = 0.8A sin(kx + \omega t + 30^{\circ} + 180^{\circ})$$
Hence option B.
Question 10
1 / -0

Vibrations of period 0.25 s propagate along a straight line at a velocity of 48 cm/s. One second after the emergence of vibrations at the initial point, displacement of the point, 47 cm from it is found to be 3 cm. Then,

A
amplitude of vibrations is 6 cm.

B
amplitude of vibrations is $$3 \sqrt{2} cm.$$

C
amplitude of vibrations is 3 cm.

D
None of the above

Solution

The wavelength of the wave can be found by using $$\dfrac{\lambda}{T}=v$$
$$\implies \lambda=vT=48cm/s\times 0.25s=12cm$$
Four full wavelengths complete at a distance of 48cm.
Thus a point 47cm lag by a phase difference of $$\dfrac{2\pi}{\lambda}(48cm-47cm)=\dfrac{\pi}{6}$$
Let the amplitude of vibrations be $$A$$.
Thus the displacement at the given point=$$Asin(\dfrac{\pi}{6})=\dfrac{A}{2}=3cm$$
$$\implies A=6cm$$
Thus correct answer is option A.

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Waves Test - 31

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Waves Test - 31

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SHARING IS CARING

Question 1

$$1$$

$$2$$

$$8$$

$$3$$

Solution

Question 2

1

2

8

4

Solution

Question 3

$$2 Hz$$

$$6 Hz$$

$$8 Hz$$

$$10 Hz$$

Solution

Question 4

$$\displaystyle \frac{\pi}{2}$$

$$\pi$$

$$2 \pi$$

$$\displaystyle \frac{\pi}{4}$$

Solution

Question 5

$$\displaystyle \frac{\pi}{4}$$

$$\displaystyle \frac{\pi}{2}$$

$$\pi$$

$$2 \pi$$

Solution

Question 6

4A

3A

2A

A

Solution

Question 7

$$3$$

$$5$$

$$2$$

$$1$$

Solution

Question 8

7 mm

10 mm

5 mm

4.8 mm

Solution

Question 9

$$Y\, =\,0.8 \,A\, sin\,(kx\,-\,\omega\,t\,+\,30^{\circ}\,+\,180^{\circ})$$

$$Y\, =\,0.8 \,A\, sin\,(kx\,+\,\omega\,t\,+\,30^{\circ}\,+\,180^{\circ})$$

$$Y\, =\,0.8 \,A\, sin\,(kx\,-\,\omega\,t\,-\,30^{\circ})$$

$$Y\, =\,0.8 \,A\, sin\,(kx\,-\,\omega\,t\,+\,30^{\circ})$$

Solution

Question 10

amplitude of vibrations is 6 cm.

amplitude of vibrations is $$3 \sqrt{2} cm.$$

amplitude of vibrations is 3 cm.

None of the above

Solution

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