Waves Test

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Waves Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

An oscilloscope is basically designed to convert .................... .

A
Visual signals to electrical signals

B
Sound signals to electrical signals

C
Electrical signals to visual signals

D
Sound signals to visual signals

Solution

Solution
Oscilloscope helps us to display and test variations in voltage signals with time in graph form.
Therefore it converts electrical signal to visual signal.
The correct option is C
Question 2
1 / -0

In a stationary wave,

A
Phase of all vibrating particles is same.

B
Phase of different particles is different.

C
Phase of particles between the two nodes is same.

D
None of these

Solution

Solution
A stationary wave is produced when two waves of same wavelength and amplitude but moving in opposite direction superimpose.
Particles at node are stationary. particles at antinodes move with high velocity
All those particles between two nodes vibrates with same phase
The correct option is C
Question 3
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Consider a function $$y = 10 \sin^{2} (100\pi t + 5 \pi z)$$ where $$y, z$$ are in $$cm$$ and $$t$$ is $$second. $$

A
the function represents a travelling, periodic wave propagating in (-z) direction with speed $$20 m/s.$$

B
the function does not represent a travelling wave.

C
the amplitude of the wave in $$5 cm.$$

D
the amplitude of the wave is $$10 cm.$$

Solution

$$\sin^2\theta = \dfrac{1}{2}(1-\cos(2\theta))$$.
So,
$$y=5[1-\cos(200\pi t+10 \pi z)]$$.
The above equation represents a plane progressive wave along -z axis with amplitude $$5 cm$$.
$$\omega = 200 \pi$$ and $$k=10 \pi$$
So velocity $$v = \dfrac{\omega}{k}=20cm/s$$
Question 4
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A string 1m long is drawn by a 300Hz vibrator attached to its end. The string vibrates in 3 segments. The speed of transverse waves in the string is equal to

A
100 m/s

B
200 m/s

C
300 m/s

D
400 m/s

Solution

Wavelength, $$ \lambda $$ $$= 2L$$.
Where, $$L$$ is the length of the segment or number of segments.
Velocity, $$v =$$ $$ \lambda f$$ $$=300\times \dfrac{2}{3} = 200\ m/s $$
Question 5
1 / -0

769Hz longitudinal wave in air has a speed of 344m/s. At a particular instant, what is the phase difference (in degrees) between two points 5.0 cm apart?

A
30

B
40

C
45

D
60

Solution

Given that,

The frequency of wave is $$\nu = 769 Hz$$

The speed of wave is $$v = 344 ms^{-1}$$

The wavelength of wave is given by

$$\lambda = \dfrac{v}{\nu}$$

$$\lambda = \dfrac{344}{769}$$

$$\lambda = 0.447 m = 44.7 cm$$

Hence, the phase difference between two points 5.0 cm apart is

$$\dfrac{5}{44.7} \times 360^\circ = 0.1118 \times 360^\circ = 40.24^\circ \approx40^\circ$$
Question 6
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A sound wave travels with a speed of $$330 ms^{-1}$$ in air. If the wavelength of the wave is 3.3m, then the frequency of the wave is ______.

A
50 Hz

B
100 Hz

C
150 Hz

D
200 Hz

Solution

since $$ v= f \times \lambda$$
$$ 330= f \times 3.3 $$
$$ f = 100$$ Hz
so the answer is B.
Question 7
1 / -0

A composition string is made up by joining two strings of different masses per unit length $$\longrightarrow \mu $$ and $$4\mu.$$ The composite string is under the same tension. A transverse wave pulse : $$Y=(6 mm) \sin (5t+40x)$$, where '$$t$$' is in seconds and '$$x$$' in meters, is sent along the lighter string towards the joint. The joint is at $$x=0.$$ The equation of the wave pulse reflected from the joint is

A
$$(2 mm) \sin(5t-40x)$$

B
$$(4 mm) \sin(40x-5t)$$

C
$$- (2 mm) \sin(5t-40x)$$

D
$$(2 mm) \sin(5t-10x)$$

Solution

Since the wave is travelling from low to more dense medium, thus there will be inversion of the reflected wave . Since it travels back from origin, thus $$40x$$ will become $$-40x.$$ Also, amplitude of reflected wave is given by:
$${ A }_{ r }=\dfrac { Z_{ 1 }-{ Z }_{ 2 } }{ { Z }_{ 1 }+Z_{ 2 } } A.\quad Z=\mu c=\mu \sqrt { \dfrac { T }{ \mu } } =\sqrt { \mu T } \\ Thus\quad { A }_{ r }=\dfrac { \sqrt { \mu T } -\sqrt { 4\mu T } }{ \sqrt { \mu T } +\sqrt { 4\mu T } } A=\dfrac { -1 }{ 3 } A\\ $$
Thus it become $$-6/3=-2$$.
Question 8
1 / -0

A pulse shown in the figure is reflected from the rigid wall A and then from free end B. The shape of the string after these 2 reflection will be :

A

B

C

D
Solution
Explanation :
Phase change because of reflection from an inflexible limit $$(\mathrm{A})=\pi$$.
Phasechange because of reflection from a more uncommon limit $$(\mathrm{B})=0$$.
Total phase change due to two reflections $$=\pi$$.
Therefore, The correct option is 'A'
Question 9
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A particle is executing SHM of amplitude $$A$$ about the mean position $$x=0$$. Which of the following cannot be a possible phase difference between the positions of the particle at $$x=+{A}/{2}$$ and $$x=-{A}/{\sqrt{2}}$$.

A
$$75^\circ$$

B
$$165^\circ$$

C
$$135^\circ$$

D
$$195^\circ$$

Solution

sine wave starts from zero,but for sin 90 degree its $$ \dfrac{1}{2}$$ whereas for sin 135 degree its $$ \dfrac{1}{ \sqrt{2} }$$ hence for sin 135 degree its value will be $$ \dfrac{1}{ \sqrt{2} }$$
therefore option (c) is correct
Question 10
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An ultrasonic source emits sound of frequency 220 kHz in air. If this sound meets a water surface, what is the wavelength of the transmitted sound? (At the atmospheric temperature, speed of sound in air $$=352 m s^{-1} and in water = 1.496 m s^{-1}$$)

A
$$5.8 \times 10^{-3} m$$

B
$$6.8 \times 10^{-3} m$$

C
$$7.8 \times 10^{-3} m$$

D
$$8.8 \times 10^{-3} m$$

Solution

Here, $$v=220 kHz = 220 \times 10^3$$
$$=2.2 \times 10^5 Hz$$;
Speed of sound in air, $$v_a = 352 m s^{-1}$$
Speed of sound in water, $$v_w = 1.496 m s^{-1}$$

The transmitted sound : The transmitted ultrasonic sound travels in water. If $$\lambda_w$$ is wavelength of ultrasonic sound in water, then
$$\lambda_w = \displaystyle \frac{v_w}{v} = \frac{1,496}{2.2 \times 10^5} = 6.8 \times 10^{-3} m$$