Waves Test

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Waves Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

Mark the incorrect statement

A
During motion of waves, energy is transferred from one point to another

B
All vibrating bodies produce sound

C
A transverse wave consists of crests and troughs

D
Transverse wave can propagate through any medium

Solution

Transverse waves can travel through solids and over the surface of liquids, but not through gases.
Question 2
1 / -0

'S' waves are

A
Longitudinal

B
Electromagnetic

C
Ultrasonic

D
Transverse

Solution

The S waves moving through the planet make the medium to move perpendicular to the direction of the propagation of the wave. Therefore. the S waves are regarded as the transverse waves.
Question 3
1 / -0

For the travelling harmonic wave $$y(x,t)=2.0 cos $$ $$ 2\pi $$ (10t-0.0080 x+0.35 ) where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of $$x$$

A
$$x=4 m,\ \ \Delta\phi=6.4π \ rad $$

B
$$0.5 m,\ \ \ \ \ \Delta\phi=0.6π \, rad $$

C
$$ \displaystyle \lambda /2 ,\ \ \ \ \ \ \ \Delta\phi= .6π \ rad$$

D
$$ \displaystyle 3\lambda /4,\ \ \ \ \ \Delta\phi= 2.5π \ rad .$$

Solution

Equation for a travelling harmonic wave is given as:
$$y(x, t)=2.0\,cos\,2\pi(10t-0.0080x+0.35)$$
$$=2.0\,cos(20\pi t-0.016\pi x+0.70\pi)$$
Where,
Propagation constant, $$k = 0.0160\pi$$
Amplitude, $$a=2\,cm$$
Angular frequency, $$\omega =20\pi\,rad/s$$
Phase difference is given by the relation:
$$\phi =kx=2\pi/\lambda$$

(a) For $$\Delta x=4m= 400 cm$$
$$\Delta \phi = 0.016\pi\times 400=6.4\pi\, rad$$

(b) For $$\Delta x=0.5 m = 50 cm$$
$$\Delta \phi = 0.016\pi \times 50 = 0.8\pi\, rad$$

(c) For $$\Delta x=\lambda/2$$
$$\Delta \phi=2\pi/\lambda \times \lambda/2=\pi\, rad$$

(d) For $$\Delta x=3\lambda/4$$
$$\Delta \phi=2\pi/\lambda \times 3\lambda/4=1.5\pi\, rad$$.
Question 4
1 / -0

The displacement of a particle performing simple harmonic motion is given by, $$x=8\sin { \omega t } +6\cos { \omega t } $$, where distance is in $$cm$$ and time is in second. The amplitude of motion is:

A
$$10 cm$$

B
$$2 cm$$

C
$$14 cm$$

D
$$3.5 cm$$

Solution

A harmonic oscillation of constant amplitude and of single frequency is called simple harmonic oscillation.
Here, $$x=8\sin { \omega t } +6\cos { \omega t } $$
So, $${ a }_{ 1 }=8cm$$ and $${ a }_{ 2 }=6cm$$
$$\therefore$$ Amplitude of motion
$$A=\sqrt { { a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 } } $$
$$=\sqrt { { 8 }^{ 2 }+{ 6 }^{ 2 } } $$
$$=\sqrt { 64+36 } =\sqrt { 100 } $$
$$=10cm$$
Question 5
1 / -0

A particle is vibrating simple harmonically with an amplitude $$a$$. The displacement of the particle when its energy is half kinetic and half potential.

A
$$\cfrac{a}{2}$$

B
$$\cfrac{a}{\sqrt{2}}$$

C
$$\cfrac{a}{4}$$

D
zero

Solution

Kinetic energy$$=$$ Potential energy
$$\Rightarrow$$ $$\cfrac{1}{2}m{\omega}^{2}({a}^{2}-{y}^{2})=\cfrac{1}{2}m{\omega}^{2}{y}^{2}$$
or $${a}^{2}-{y}^{2}={y}^{2}$$
or $$2{y}^{2}={a}^{2}$$
or $${y}^{2}=\cfrac{{a}^{2}}{2}$$
or $$y=\cfrac{a}{\sqrt {2}}$$
Question 6
1 / -0

Nowadays, a lot of technological advancement has occurred and attempts are made to make system more efficient. Sound from High Tech phones do not exhibit reverberation. What could be the reason?

A
There is no reverberation from the actual source

B
Reverberations are negligible

C
Techniques are used to remove echoes

D
The above statement is false

Solution

Echoes are copies of sound produced by reflections. These copies can be detected and removed from a recording.
Question 7
1 / -0

The displacement equation of a simple harmonic oscillator is given by
$$A\sin { \omega t-B\cos { \omega t } } $$
The amplitude of the oscillator will be :

A
$$A-B$$

B
$$A+B$$

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

D
$$({ A }^{ 2 }+{ B }^{ 2 })$$

Solution

Displacement equation
$$y=A\sin { \omega t } -B\cos { \omega t } $$
Let $$A=a\cos { \theta } $$ and $$N=a\sin { \theta } $$
So, $${A}^{2}+{B}^{2}={a}^{2}$$
$$\Rightarrow$$ $$a=\sqrt {{A}^{2}+{B}^{2}}$$
Then $$y=a\cos { \theta } \sin { \omega t } -a\sin { \theta } \cos { \omega t } $$
$$y=a\sin { \left( \omega t-\theta \right) } $$
which is the equation of simple harmonic oscillator
The amplitude of the oscillator
$$a=\sqrt {{A}^{2}+{B}^{2}}$$
Question 8
1 / -0

Two strings with mass per unit length of $$25\ g/cm$$ and $$9\ g/cm$$ are joined together in series. The reflection coefficient for the vibration waves are

A
$$\dfrac {9}{25}$$

B
$$\dfrac {3}{5}$$

C
$$\dfrac {1}{16}$$

D
$$\dfrac {9}{16}$$

Solution

Let $$I_{r}$$ and $$I_{i}$$ represent the intensities of reflected and incident waves respectively, then
$$\dfrac {I_{r}}{I_{i}} = \left (\dfrac {\mu - 1}{\mu + 1}\right )^{2}$$
where $$\mu = \dfrac {v_{1}}{v_{2}}$$
or $$v = \dfrac {\sqrt {\dfrac {T}{m_{1}}}}{\sqrt {\dfrac {T}{m_{2}}}} = \sqrt {\dfrac {m_{2}}{m_{1}}} = \sqrt {\dfrac {25}{9}} = \dfrac {5}{3}$$
$$\therefore \dfrac {I_{r}}{I_{i}} =\left [\dfrac {\left (\dfrac {5}{3}\right ) - 1}{\left (\dfrac {5}{3}\right ) + 1}\right ]^{2} = \dfrac {1}{16}$$
Question 9
1 / -0

A long spring whose one end is fixed is stretched from the other end and the left longitudinal waves of frequency $$500 Hz$$ are produced. If the velocity of wave is $$250 m/s$$. Find the distance between two consecutive compression and rarefaction :

A
$$1.25 m$$

B
$$0.5m$$

C
$$2.5m$$

D
$$2m$$

Solution

$$f=500 Hz$$
$$v= 250 m/s$$
$$v=f\times \lambda$$
or, $$250=500 \times \lambda $$
or, $$\lambda =0.5m$$
$$\therefore$$ distance between 2 consecutive compression = wavelength $$=0.5m$$
Question 10
1 / -0

Mark the correct statement:
I) Directions in which the sound is incident and is reflected make equal angles with the normal to the reflecting surface
II) Reflection of sound occurs from a polished surface
III) Reflection of sound occurs from a rough surface

A
I, II

B
I, III

C
II, III

D
All

Solution

Sound bounces off a solid or a liquid like a rubber ball bounce off a wall. Like light, the sound gets reflected at the surface of a solid or liquid and follows the same laws of reflection.

The directions in which the sound is incident and is reflected make equal angles with the normal to the reflecting surface, and the three are in the same plane.

An obstacle of large size which may be polished or rough is needed for the reflection of sound waves.

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Waves Test - 36

Result

Waves Test - 36

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SHARING IS CARING

Question 1

During motion of waves, energy is transferred from one point to another

All vibrating bodies produce sound

A transverse wave consists of crests and troughs

Transverse wave can propagate through any medium

Solution

Question 2

Longitudinal

Electromagnetic

Ultrasonic

Transverse

Solution

Question 3

$$x=4 m,\ \ \Delta\phi=6.4π \ rad $$

$$0.5 m,\ \ \ \ \ \Delta\phi=0.6π \, rad $$

$$ \displaystyle \lambda /2 ,\ \ \ \ \ \ \ \Delta\phi= .6π \ rad$$

$$ \displaystyle 3\lambda /4,\ \ \ \ \ \Delta\phi= 2.5π \ rad .$$

Solution

Question 4

$$10 cm$$

$$2 cm$$

$$14 cm$$

$$3.5 cm$$

Solution

Question 5

$$\cfrac{a}{2}$$

$$\cfrac{a}{\sqrt{2}}$$

$$\cfrac{a}{4}$$

zero

Solution

Question 6

There is no reverberation from the actual source

Reverberations are negligible

Techniques are used to remove echoes

The above statement is false

Solution

Question 7

$$A-B$$

$$A+B$$

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

$$({ A }^{ 2 }+{ B }^{ 2 })$$

Solution

Question 8

$$\dfrac {9}{25}$$

$$\dfrac {3}{5}$$

$$\dfrac {1}{16}$$

$$\dfrac {9}{16}$$

Solution

Question 9

$$1.25 m$$

$$0.5m$$

$$2.5m$$

$$2m$$

Solution

Question 10

I, II

I, III

II, III

All

Solution

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