Waves Test

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Waves Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

The minimum phase difference between two simple harmonic oscillations,
$${ y }_{ 1 }=\dfrac { 1 }{ 2 } \sin { \omega t } +\dfrac { \sqrt { 3 } }{ 2 } \cos { \omega t } $$
$${ y }_{ 2 }=\sin { \omega t } +\cos { \omega t } $$

A
$$\dfrac { 7\pi }{ 12 } $$

B
$$\dfrac { \pi }{ 12 } $$

C
$$-\dfrac { \pi }{ 3 } $$

D
$$\dfrac { \pi }{ 6 } $$

Solution

Phase difference between any two particles in a wave determines lack of harmony in the vibrating state of two particles ie, how far one particle leads the other or lags behind the other.
Here, $${ y }_{ 1 }=\dfrac { 1 }{ 2 } \sin { \omega t } +\dfrac { \sqrt { 3 } }{ 2 } \cos { \omega t } $$
$$=\cos { \dfrac { \pi }{ 3 } } \sin { \omega t } +\sin { \dfrac { \pi }{ 2 } } \cos { \omega t } $$
$$\therefore { y }_{ 1 }=\sin { \left( \omega t+\dfrac { \pi }{ 3 } \right) } $$
Similarly, $${ y }_{ 2 }=\sqrt { 2 } \sin { \left( \omega t+\dfrac { \pi }{ 4 } \right) } $$
$$\therefore $$ Phase difference $$\Delta \phi =\dfrac { \pi }{ 3 } -\dfrac { \pi }{ 4 } $$
$$=\dfrac { \pi }{ 12 } $$
Question 2
1 / -0

The equation of sound wave is $$y=0.0015\;sin(62.4x+316t)$$. Find the wavelength of this wave :

A
$$0.2$$ unit

B
$$0.1$$ unit

C
$$0.3$$ unit

D
none of these

Solution

The equation for sound wave is given as $$y = 0.0015 sin(62.4x + 316 t)$$
On comparing with $$y = A sin (kx+wt)$$, we get $$k = 62.4$$
Let the wavelength of the wave be $$\lambda$$.
$$\therefore$$ $$\dfrac{2\pi}{\lambda} = 62.4$$ $$\implies \lambda = 0.1$$ unit
Question 3
1 / -0

The frequency of radio waves corresponding to a wavelength of 10 m is

A
$$3\times 10^7\,Hz$$

B
$$3.3\times 10^8\,Hz$$

C
$$3\times 10^9\,Hz$$

D
$$3\times 10^{-7}\,Hz$$

Solution

For radio waves, speed $$v=c=3 \times 10^8\ m/s$$
Given, $$\lambda=10\ m$$
$$v=f\lambda$$
$$f=3 \times 10^7\ Hz$$
Question 4
1 / -0

Let, the velocity of a wave be $$V$$, time period $$T$$, frequency $$\gamma$$ and wavelength $$\lambda$$. Mark the correct relation between these three.

A
$$VT = \dfrac{1}{\lambda}$$

B
$$VT = \lambda$$

C
$$\dfrac{V}{T} = \lambda$$

D
$$\dfrac{T}{V} = \lambda$$

Solution

$$\lambda =$$ Distance travelled by the wave in one time period i.e., in $$T$$ second,
$$=$$ wave velocity $$\times$$ Time period
$$= V \times T$$
$$VT = \lambda$$
Question 5
1 / -0

The phase difference between two points is $$\pi/3$$. If the frequency of wave is 50 Hz, then what is the distance between two points? (given v = 330 m/s)

A
2.2 m

B
1.1 m

C
0.6 m

D
1.7 m

Solution

Phase difference $$= \dfrac{2\pi}{\lambda}\times$$path difference.
The phase difference between any two particles in a wave determines lack of harmony in the vibrating state of two particles ie, how far one particle leads the other or lags behind the other.
From relation
$$\Delta \phi =\dfrac{2\pi}{\lambda}\times \delta x$$

$$\Rightarrow \Delta x = \dfrac{\lambda}{2\pi}\times \Delta \phi$$ ...(i)

Also, $$\lambda = \dfrac{v}{n}$$ ...(ii)
Now, from Eqs. (i) and (ii), we get
$$\Delta x=\dfrac{v}{2\pi n}\times \Delta \phi$$

$$\Rightarrow \Delta x=\dfrac{330}{2\pi\times 50}\times \dfrac{pi}{3}$$
or $$\Delta x = 1.1m$$
Question 6
1 / -0

The speed of sound in air is $$320m/s$$. The fundamental frequency of an open pipe $$50cm$$ long will be

A
$$320Hz$$

B
$$160Hz$$

C
$$640Hz$$

D
$$960Hz$$

Solution

The frequency of open pipe$$=\cfrac{v}{2l}$$
$$\Rightarrow$$ $$n=\cfrac{320}{2\times 0.5}$$
or $$n=\cfrac{320}{1.0}=320Hz$$
Question 7
1 / -0

A wave is travelling in a medium with frequency $$f = {10}^{14} Hz$$ and speed $$v = 100 {m}/{s}$$, Find out its wavelength?

A
$${10}^{-16} m$$

B
$${10}^{-12} m$$

C
$${10}^{10} m$$

D
$${10}^{12} m$$

E
$${10}^{16} m$$

Solution

Wavelength of the wave $$\lambda = \dfrac{v}{f} = \dfrac{100}{10^{14}} = 10^{-12}$$ $$m$$
Question 8
1 / -0

Find out the frequency of orange light whose wavelength is $$6 \times {10}^{-7} m$$. The speed of light is $$3 \times {10}^{8} {m}/{s}$$.

A
$$2 \times {10}^{15} Hz$$

B
$$2 \times {10}^{-15} Hz$$

C
$$5 \times {10}^{14} Hz$$

D
$$5 \times {10}^{-14} Hz$$

E
$$2 \times {10}^{14} Hz$$

Solution

Given : $$\lambda = 6\times 10^{-7}$$ $$m$$ $$v = 3\times 10^8$$ $$m/s$$
Frequency of light $$\nu =\dfrac{v}{\lambda} = \dfrac{3\times 10^8}{6\times 10^{-7}} =5 \times 10^{14}$$ $$Hz$$
Question 9
1 / -0

Which wave parameter can be found if the wavelength & the frequency is known to us?

A
speed

B
frequency

C
amplitude

D
wavelength

E
period

Solution

Speed of the wave is equal to wavelength of the wave times its frequency.
$$\therefore$$ Speed $$v = \lambda\times \nu$$
Thus option A is correct.
Question 10
1 / -0

A particle executes simple harmonic motion with amplitude $$A$$. The distance moved by the particle in one oscillation is:

A
Zero

B
$$A$$

C
$$2A$$

D
$$4A$$

Solution

Let the distance between the two extreme positions be $$d$$.
Amplitude is $$A$$
$$\implies d=2A$$
Distance travelled by the particle in one oscillation is $$2d$$ (from one extreme to the second and back from the second extreme to the first).
Hence, total distance $$=2d=4A$$

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Re-Attempt Weekly Quiz Competition

Waves Test - 15

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Waves Test - 15

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SHARING IS CARING

Question 1

$$\dfrac { 7\pi }{ 12 } $$

$$\dfrac { \pi }{ 12 } $$

$$-\dfrac { \pi }{ 3 } $$

$$\dfrac { \pi }{ 6 } $$

Solution

Question 2

$$0.2$$ unit

$$0.1$$ unit

$$0.3$$ unit

none of these

Solution

Question 3

$$3\times 10^7\,Hz$$

$$3.3\times 10^8\,Hz$$

$$3\times 10^9\,Hz$$

$$3\times 10^{-7}\,Hz$$

Solution

Question 4

$$VT = \dfrac{1}{\lambda}$$

$$VT = \lambda$$

$$\dfrac{V}{T} = \lambda$$

$$\dfrac{T}{V} = \lambda$$

Solution

Question 5

2.2 m

1.1 m

0.6 m

1.7 m

Solution

Question 6

$$320Hz$$

$$160Hz$$

$$640Hz$$

$$960Hz$$

Solution

Question 7

$${10}^{-16} m$$

$${10}^{-12} m$$

$${10}^{10} m$$

$${10}^{12} m$$

$${10}^{16} m$$

Solution

Question 8

$$2 \times {10}^{15} Hz$$

$$2 \times {10}^{-15} Hz$$

$$5 \times {10}^{14} Hz$$

$$5 \times {10}^{-14} Hz$$

$$2 \times {10}^{14} Hz$$

Solution

Question 9

speed

frequency

amplitude

wavelength

period

Solution

Question 10

Zero

$$A$$

$$2A$$

$$4A$$

Solution

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