Waves Test

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Waves Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

A wave is propagating along a string with a speed of $$320 {cm}/{s}$$ as shown in the figure above.
Find out the amplitude of the given travelling wave ?

A
$$1 cm$$

B
$$2 cm$$

C
$$4 cm$$

D
$$8 cm$$

E
$$16 cm$$

Solution

Amplitude of the wave is equal to height of the crest or depth of the trough.
From figure, magnitude of height of crest or depth of trough is $$1$$ $$cm$$
Thus amplitude of wave is $$1$$ $$cm$$
Question 2
1 / -0

The picture below shows too "snapshots" of the same spring-mass system. In the snapshot on the left, the spring is completely uncompressed and completely unstretched. In the snapshot on the right, the spring is stretched as much as it will be during the oscillating motion.
The time it takes to move from the one position to the other is shown, as is the vertical distance.
With the information in this diagram, what two quantities could be determined?

A
Spring constant and amplitude of oscillation

B
Period and amplitude of oscillation

C
Spring constant and period of oscillation

D
Spring constant and amount of mass hanging

E
Amplitude of oscillation and amount of mass hanging

Solution

The left side diagram (unstretched position) shows the mean position of spring. As given, the spring is stretched as much as it will be during motion, therefore $$0.080m$$ , will be amplitude of oscillation.
Now, spring takes $$1.50s$$ to reach its extreme position from mean position therefore time period of oscillation will be the time taken by mass to reach again in its initial position (after one oscillation) .i.e. $$T=4\times1.50=6s$$ .
And time period, $$T=2\pi\sqrt{m/k}$$,
Hence, mass and spring constant cannot be calculated by time period.
Question 3
1 / -0

Two waves interfere with each other then which of the following statement is true?

A
They always create a larger wave than either of the individual waves

B
The larger wave always cancels the smaller wave

C
The displacements of the individual waves are added at each point

D
An antinode is always created

E
An node is always created

Solution

The principle of superposition of waves states that the displacement amplitude of two waves get added upon their interaction according to their phase difference.
If the waves are in-phase, the amplitudes gets added simply.
If the waves are completely out of phase, the amplitudes are subtracted to give the resultant amplitude.
Question 4
1 / -0

During interference of waves, the amplitude of the resulting wave can be found at any position using the principle of:

A
superposition

B
interference

C
diffraction

D
none of these

Solution

The principle of superposition of waves states that the displacement amplitude of two waves get added upon their interaction according to their phase difference.
If the waves are in-phase, the amplitudes gets added simply.
If the waves are completely out of phase, the amplitudes are subtracted to give the resultant amplitude.
Question 5
1 / -0

Identify which one of the following can be represented by the product of the other two?
I. Speed
II. Wavelength
III. Frequency

A
I

B
II

C
III

D
Any one is the product of the other two

E
None of the three is the product of the other two

Solution

Let the speed, wavelength and frequency of the wave to be represented by $$v$$, $$\lambda$$ and $$\nu$$ respectively.
$$\therefore$$ Speed $$v = \lambda \times \nu$$
Hence option A is correct.
Question 6
1 / -0

A wave of frequency 500 Hz has a velocity of 360 m $$s^{-1}$$. Calculate the distance between two points that are $$60^0$$ out of phase.

A
12 cm

B
18 cm

C
24 cm

D
32 cm

Solution

frequency $$\left( \nu \right) =500{ H }_{ z }$$
velocity $$\left( v \right) =360m/s$$
$$\therefore $$ wavelength $$=$$ velocity/frequency
$$=\cfrac { 360 }{ 500 } =0.72m=72cm$$
$$\therefore \quad \lambda =72cm$$
$${ 60 }^{ ° }$$ out of phase implies that the wavelength between two points is
$$\cfrac { \lambda }{ \dfrac { { 360 }^{ ° } }{ { 60 }^{ ° } } } =\dfrac { \lambda }{ 6 }$$
$$\therefore \quad { 360 }^{ ° }=\lambda $$
$${ 1 }^{ ° }=\cfrac { \lambda }{ 360 } $$
$${ 60 }^{ ° }=\cfrac { \lambda \times 60 }{ 360 } =\cfrac { \lambda }{ 6 } =\cfrac { 72 }{ 6 } =12cm$$
$$\therefore $$ The distance between point $$A$$ & $$B$$ is $$12cm$$ if they have a phase difference of $${ 60 }^{ ° }$$
$$\therefore $$ Option (A) is the correct answer.
Question 7
1 / -0

The equation, $$y = A \sin [2(vt- x)]$$ represent a plane progressive harmonic wave proceeding along the positive X-direction. The equation, $$y = A \sin[2(x -vt)]$$ represents

A
a plane progressive harmonic wave proceeding along the negative X-direction

B
a plane progressive harmonic wave with a phase difference of proceeding along the negative X-direction

C
a plane progressive harmonic wave with a phase difference of proceeding along the positive X-direction

D
none of the above
Question 8
1 / -0

What will be the wave velocity, if the radar gives $$54$$ waves per min and wavelength of the given wave is $$10 m$$?

A
$$4 m {s}^{-1}$$

B
$$6 m {s}^{-1}$$

C
$$9 m {s}^{-1}$$

D
$$5 m {s}^{-1}$$

Solution

If the radar gives $$54$$ waves per minute and the wavelength$$\lambda$$ is $$10m$$, then it means that the wave travels a distance of $$54\times10$$metres in one minute.
Since the SI unit of velocity(v) m/s, $$v=\cfrac{54\times10}{60}=$$ 9m/s
Question 9
1 / -0

When a transverse wave on a string is reflected from the free end, the phase change produced is ___________.

A
Zero rad

B
$$\dfrac { \pi }{ 2 } $$ rad

C
$$\dfrac { 3\pi }{ 4 } $$ rad

D
$$\pi$$ rad

Solution

For a transverse wave, a phase change of $$\pi$$ occurs when it is reflected from a denser medium.
When reflected from a free end, however, there is no change of phase.
Hence the correct answer is option A.
Question 10
1 / -0

The distance between two consecutive crests in a wave train produced on a string is $$5 cm$$. If two complete waves pass through any point per second, the velocity of the wave is

A
$$10 cm {s}^{-1}$$

B
$$2.5 cm {s}^{-1}$$

C
$$5 cm {s}^{-1}$$

D
$$15 cm {s}^{-1}$$

Solution

Hint:-Use the formula of the velocity of the wave.

Step 1: Note the given values.
Wavelength of wave,$$\lambda$$=distance between two consecutive crests=5cm
Two complete waves pass through any point per second so the frequency is
$$f= \dfrac {\text {no.of cycles}}{\text{time}}=\dfrac21 =2s^{-1}$$

Step 2: Calculate velocity of wave.
$$v=f\lambda$$
$$v=2\times5= 10cms^{-1}$$

$$\textbf{Hence option A correct}$$

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Waves Test - 17

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Waves Test - 17

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Question 1

$$1 cm$$

$$2 cm$$

$$4 cm$$

$$8 cm$$

$$16 cm$$

Solution

Question 2

Spring constant and amplitude of oscillation

Period and amplitude of oscillation

Spring constant and period of oscillation

Spring constant and amount of mass hanging

Amplitude of oscillation and amount of mass hanging

Solution

Question 3

They always create a larger wave than either of the individual waves

The larger wave always cancels the smaller wave

The displacements of the individual waves are added at each point

An antinode is always created

An node is always created

Solution

Question 4

superposition

interference

diffraction

none of these

Solution

Question 5

I

II

III

Any one is the product of the other two

None of the three is the product of the other two

Solution

Question 6

12 cm

18 cm

24 cm

32 cm

Solution

Question 7

a plane progressive harmonic wave proceeding along the negative X-direction

a plane progressive harmonic wave with a phase difference of proceeding along the negative X-direction

a plane progressive harmonic wave with a phase difference of proceeding along the positive X-direction

none of the above

Question 8

$$4 m {s}^{-1}$$

$$6 m {s}^{-1}$$

$$9 m {s}^{-1}$$

$$5 m {s}^{-1}$$

Solution

Question 9

Zero rad

$$\dfrac { \pi }{ 2 } $$ rad

$$\dfrac { 3\pi }{ 4 } $$ rad

$$\pi$$ rad

Solution

Question 10

$$10 cm {s}^{-1}$$

$$2.5 cm {s}^{-1}$$

$$5 cm {s}^{-1}$$

$$15 cm {s}^{-1}$$

Solution

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