Waves Test

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Waves Test - 62

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Weekly Quiz Competition

Question 1
1 / -0

For a wave displacement amplitude is $$10^-4 m$$,density of air 1.3 kg $$m^-4$$,velocity in air 340 $$ms^-4$$ are frequency is 2000 Hz.The intensity of wave is

A
$$5.3\times 10^-4 Wm^-4$$

B
$$5.3\times 10^4 Wm^-4$$

C
$$3.5\times 10^-4 Wm^4$$

D
none of these

Solution

$$\begin{array}{l} From\, the\, question \\ A={ 10^{ -8 } }m \\ \rho =1.3\, kg/{ m^{ 3 } } \\ v=340\, m/s \\ v=2000\, Hz \\ The\, \, { { int } }ensity\, of\, wave\, \, is\, given\, by\, the\, following\, relation \\ I=\frac { 1 }{ 2 } \rho { A^{ 2 } }{ \omega ^{ 2 } }c\, \, \, \, \, \, \, \, \, \, ---\left( 1 \right) \\ \omega \, is\, angular\, frequency\, ,\, \omega =2\pi v \\ c\, is\, the\, speed\, of\, light \\ So,\, equation\, \left( 1 \right) \, becomes \\ I=\frac { 1 }{ 2 } \times 1.3\, \, kg/{ m^{ 3 } }\times { \left( { { { 10 }^{ -8 } }m } \right) ^{ 2 } }\times { \left( { 2\pi \times 2000\, Hz } \right) ^{ 2 } }\times 3\times { 10^{ 8 } }m/s \\ I=3.0761\, W/{ m^{ 2 } } \end{array}$$
Hence, the option $$D$$ is the correct answer.
Question 2
1 / -0

The S.H.M. of a particle is given by the equation $$y = 3 \sin \omega t + 4 \cos \omega t .$$The amplitude is

A
7

B
1

C
5

D
12

Solution
Question 3
1 / -0

Two equations of two SHM are
$$x=a \ sin(\omega t - \alpha) \ and \ y=b \ cos(\omega t - \alpha)$$
The phase difference between the two is

A
$$0^o$$

B
$$\alpha^o$$

C
$$90^o$$

D
$$180^o$$

Solution

$$\begin{array}{l} Equation\; \, for\, \, \, volatage\, \, is\; \, given\, as: \\ V={ V_{ 0 } }\cos \omega t \\ Equation\, \, for\, \, \, current\, \, is\, \, given\, as\, \, : \\ l={ l_{ .0 } }\cos \left( { \omega t-\phi } \right) \\ Where, \\ \phi =phase\, \, difference\, \, between\, \, voltage\, \, \, and\, \, \, currect \\ At\, \, time,t=0 \\ V={ V_{ 0 } }\left( { voltage\, \, is\, \, \, \max imum } \right) \\ For\, \, \omega t-\phi =0 \end{array}$$
Question 4
1 / -0

Three simple harmonic motions in the same direction having the same amplitude $$a$$ and period are superposed. IF each differs in phase from the next by $${45}^{o}$$ then

A
the resultant amplitude $$(1+\sqrt{2})a$$

B
the phase of the resultant motion relative to the first is $${90}^{o}$$

C
the energy associated with the resulting motion is $$(3+2\sqrt{2})$$ times the energy associated any single motion

D
the resulting motion is not simple harmonic

Solution

$$\begin{array}{l} Let\, simple\, harmonic\, motions\, be\, represented\, by\, \\ { y_{ 1 } }=a\sin \left( { \omega t-\frac { \pi }{ 4 } } \right) \, \, ;\, \, { y_{ 2 } }=s\, \sin \omega t\, and \\ { y_{ 3 } }=a\sin \left( { \omega t+\frac { \pi }{ 4 } } \right) \\ On\, erimpo\sin g\, resul\tan t\, SHM\, will\, be \\ y=a\left[ { \sin \left( { \omega t-\frac { \pi }{ 4 } } \right) +\sin \omega t+\sin \left( { \omega t+\frac { \pi }{ 4 } } \right) } \right] \\ =a\left[ { 2\sin \omega t\cos \frac { \pi }{ 4 } +\sin \omega t } \right] \\ =a\left[ { \sqrt { 2 } \sin \omega t+\sin \omega t } \right] \\ =a\left( { 1+\sqrt { 2 } } \right) \sin \omega t \\ resul\tan t\, amplitude\, =\left( { 1+\sqrt { 2 } } \right) a \\ Now, \\ \frac { { { E_{ { { Re } }sul\tan t } } } }{ { { E_{ Single } } } } ={ \left( { \frac { A }{ a } } \right) ^{ 2 } }={ \left( { \sqrt { 2 } +1 } \right) ^{ 2 } }=\left( { 3+2\sqrt { 2 } } \right) \\ { E_{ { { Re } }sul\tan t } }=\left( { 3+2\sqrt { 2 } } \right) { E_{ S{ { ingle } } } } \end{array}$$
Hence, the option $$C$$ is the correct answer.
Question 5
1 / -0

Four waves of the equation given by
$$y_1= 5A \ sin \ (wt - Kx + \pi/2)$$
$$y_2= 2A \ sin \ (wt - Kx + 3 \pi/2)$$
$$y_3= 6A \ sin \ (wt - Kx)$$
and $$y_4= 2A \ sin \ (wt - Kx + \pi)$$

A
the resultant wave will have amplitude 5A.

B
It will make a phase difference of $$37^o$$ with $$\gamma_1$$.

C
Its equation will be $$y=5A \ sin \ (wt + 37^o).$$

D
Its equation will be $$y=5A \ sin \ (wt -Kx + 37^o).$$
Question 6
1 / -0

A particle moving along a straight line with a constant acceleration of - 4 m per second square passes through a point on the line with a velocity of + 8 metre per second at the moment when the distance travelled by the particle in 6 second in 5 seconds

A
$$5 m/s^2$$

B
$$6 m/s^2$$

C
$$8 m/s^2$$

D
$$2 m/s^2$$

Solution

$$uy=10sin 60=53√m/s$$
$$⇒t=2uyg=2×53√10=3√s$$
$$Sx=uxt+12axt^2$$
$$1.15=5×t−12a×t^2$$
$$1.15=5×3√−32a$$
$$3a^2=5×1.73−1.15=8.65−1.15$$
$$3a^2=7.5 $$
$$⇒a=153=5m/s^2$$
Question 7
1 / -0

Equations of motion in the same direction are given by:
$$y_1 = a \ sin \ (\omega t - kx)$$
$$y_2 = a \ sin \ (\omega t - kx - \theta)$$
The amplitude of the medium particle will be:

A
$$2a\cos \frac{\theta }{2}$$

B
$$\sqrt{2}a \ cos \ \theta$$

C
$$4a \ cos \ \theta/2$$

D
$$\sqrt{2}a \ cos \ \theta/2$$

Solution

$$\begin{array}{l} The\, \, resul\tan t\, \, amplitude\, \, is\, \, given\, \, by \\ { a_{ R } }=\sqrt { { a^{ 2 } }+{ a^{ 2 } }+2\cdot a\cdot a\cos \theta } \\ =\sqrt { 2{ a^{ 2 } }\left( { 1+\cos \theta } \right) } \\ =2a\cos \frac { \theta }{ 2 } \, \, \, \, \, \, \, \, \, \, \, \, \left( { H\cos \theta =2{ { \cos }^{ 2 } }\frac { \theta }{ 2 } } \right) \end{array}$$
Hence,
option $$(A)$$ is correct answer.
Question 8
1 / -0

Equations of motion in the same direction are given by:
$$y_1 = 2a \ sin \ (\omega t - kx)$$
$$y_2 = 2a \ sin \ (\omega t - kx - \theta)$$
The amplitude of the medium particle will be:

A
$$2a\cos \frac{\theta }{2}$$

B
$$\sqrt{2}a \ cos \theta$$

C
$$4a \ cos \theta/2$$

D
$$\sqrt{2}a \ cos \theta/2$$

Solution

$$\begin{array}{l} The\, \, resul\tan t\, \, amplitude\, \, is\, \, given\, \, by \\ { a_{ R } }=\sqrt { { a^{ 2 } }+{ a^{ 2 } }+2\cdot a\cdot a\cos \theta } \\ =\sqrt { 2{ a^{ 2 } }\left( { 1+\cos \theta } \right) } \\ =2a\cos \frac { \theta }{ 2 } \, \, \, \, \, \, \, \, \, \, \, \, \left( { H\cos \theta =2{ { \cos }^{ 2 } }\frac { \theta }{ 2 } } \right) \end{array}$$
Hence,
option $$(A)$$ is correct answer.
Question 9
1 / -0

A small mass executes linear SHM about $$O$$ with amplitude $$a$$ and period $$T$$. Its displacement from $$O$$ at time $$T/S$$ after passing through $$O$$ is

A
$$a/S$$

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

C
$$a/2$$

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

Solution

$$\begin{array}{l} y=a\sin wt \\ =a\sin \frac { { 2\pi } }{ T } .\frac { T }{ 4 } \\ =a\sin \frac { \pi }{ 4 } =\frac { a }{ { \sqrt { 2 } } } \end{array}$$
Hence, the option $$D$$ is the correct answer.
Question 10
1 / -0

If the fundamental frequency of string is $$220 \mathrm { cps }$$, the frequency of fifth harmonic will be

A
$$44\mathrm { cps }$$

B
$$55 \mathrm { cps }$$

C
$$1100 \mathrm { cps }$$

D
$$440 \mathrm { cps }$$

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Re-Attempt Weekly Quiz Competition

Waves Test - 62

Result

Waves Test - 62

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SHARING IS CARING

Question 1

$$5.3\times 10^-4 Wm^-4$$

$$5.3\times 10^4 Wm^-4$$

$$3.5\times 10^-4 Wm^4$$

none of these

Solution

Question 2

7

1

5

12

Solution

Question 3

$$0^o$$

$$\alpha^o$$

$$90^o$$

$$180^o$$

Solution

Question 4

the resultant amplitude $$(1+\sqrt{2})a$$

the phase of the resultant motion relative to the first is $${90}^{o}$$

the energy associated with the resulting motion is $$(3+2\sqrt{2})$$ times the energy associated any single motion

the resulting motion is not simple harmonic

Solution

Question 5

the resultant wave will have amplitude 5A.

It will make a phase difference of $$37^o$$ with $$\gamma_1$$.

Its equation will be $$y=5A \ sin \ (wt + 37^o).$$

Its equation will be $$y=5A \ sin \ (wt -Kx + 37^o).$$

Question 6

$$5 m/s^2$$

$$6 m/s^2$$

$$8 m/s^2$$

$$2 m/s^2$$

Solution

Question 7

$$2a\cos \frac{\theta }{2}$$

$$\sqrt{2}a \ cos \ \theta$$

$$4a \ cos \ \theta/2$$

$$\sqrt{2}a \ cos \ \theta/2$$

Solution

Question 8

$$2a\cos \frac{\theta }{2}$$

$$\sqrt{2}a \ cos \theta$$

$$4a \ cos \theta/2$$

$$\sqrt{2}a \ cos \theta/2$$

Solution

Question 9

$$a/S$$

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

$$a/2$$

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

Solution

Question 10

$$44\mathrm { cps }$$

$$55 \mathrm { cps }$$

$$1100 \mathrm { cps }$$

$$440 \mathrm { cps }$$

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