Waves Test

Result

Waves Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

Two waves whose intensity are same $$I$$ move towards a point $$P$$ in same phase, then the resultant intensity at point $$P$$ will be :

A
$$4\, I$$

B
$$2\, I$$

C
$$\sqrt{2}\, I$$

D
$$\sqrt{6}\, I$$

Solution
Question 2
1 / -0

A particle performs SHM with amp $$A$$. Its speed is hetled when at $$\dfrac {2A}{3}$$ from equation. Two new amp is

A
$$3A$$

B
$$A\sqrt {3}$$

C
$$\dfrac {7A}{3}$$

D
$$\dfrac {A}{3}\sqrt {41}$$
Question 3
1 / -0

Equation of progressive wave is given by $$y=a\sin \pi { (\frac {t }{ 2}- \frac { x }{4 } ) } $$,where t is in seconds and x is in meter. Then the distance through which the wave moves in 4 seconds is (in meter)-

A
$$4m$$

B
$$2m$$

C
$$16m$$

D
$$8m$$

Solution

y=a sin$$ \pi (\frac{t}{2}-\frac{x}{4})$$
Compare this with standard equation, y=A sin 2$$\pi (\frac{t}{T}-\frac{x}{\lambda })$$
where, A= amplitude
$$\lambda $$= wavelength = 4
T= time period = 2
$$\frac{1}{T}=\frac{1}{2}= 0.5$$
f= frequency
$$f=\frac{1}{T}= 0.5$$
f= 0.5 Hz
Wave velocity, $$v=f\times \lambda $$
$$v=0.5\times 4=2 cm/s$$
Question 4
1 / -0

Two wave are represnted by the equations $$ y_1= a sin \omega t \ and \ y_2 = a cos \omega t $$ . the first wave :

A
leads the second by $$ \pi $$

B
lags the second by $$ \pi $$

C
leads the second by $$\frac { \pi}{2} $$

D
lags the second by $$\frac { \pi}{2} $$

Solution

Given,

$$y_1 = a\sin \omega t$$

$$y_2 = a\cos\omega t$$
$$ = a \sin ( \omega t + \pi/2)$$

So, $$y_1$$ lags $$y_2$$ by $$\pi /2$$.

So, the correct option will be $$(D)$$
Question 5
1 / -0

Three coherent waves having amplitudes 12 mm, 6 mm and 4 mm arrive at a given point with successive phase difference of $$\frac{\pi}{2}$$. Then the amplitude of the resultant wave is:

A
7 mm

B
10 mm

C
5 mm

D
1.8 mm

Solution

From the vector diagram,
$$(A_{net})_{13}=\sqrt{A_1^1+A_3^2+2A_1A_3cos\delta}$$

$$(A_{net})_{13}=\sqrt{(12)^2+(4)^2+2\times 12\times 4\times cos180^0}=8mm$$

Now the net amplitude,
$$A_n=\sqrt{A_{13}^2+A_2^2+2A_2A_{13}cos90^0}=\sqrt{A_{13}^2+A_2^2}$$

$$A_n=\sqrt{(6)^2+(8)^2}=10mm$$

The correct option is B.
Question 6
1 / -0

The phase difference between two points separated 0.8 m in a wave of frequency 120 HZ is 0.5 $$\pi $$ the value velocity is

A
144 $${ ms }^{ -1 }$$

B
384 $${ ms }^{ -1 }$$

C
256 $${ ms }^{ -1 }$$

D
720 $${ ms }^{ -1 }$$

Solution
Question 7
1 / -0

The displacement of a particle along the x axis is given by $$x=a{ sin }^{ 2 }\omega t.$$ The motion of the particle corresponds to

A
simple harmonic motion of frequency $$\omega /\pi $$

B
simple harmonic motion of frequency$$3\omega /2\pi $$

C
non simple harmonic motion

D
simple harmonic motion of frequency $$\omega /2\pi $$

Solution
Question 8
1 / -0

Displacement of particle along x-axis by $$x=a\sin^{2}{\omega t}$$. Its motion is SHM of frequency :

A
$$\omega /\pi$$

B
$$3\omega /2\pi$$

C
$$\omega /2\pi$$

D
$$not\ SHM$$

Solution

We have,

$$x= a\sin^2 \omega t$$

$$\Rightarrow \dfrac{dx}{dt} = 2 \omega (\sin \omega t) ( \cos \omega t)$$

$$\Rightarrow \dfrac{d^2x}{dt^2} = 2 \omega (\cos ^2 \omega t - \sin^2 \omega t)$$

$$\Rightarrow a = 2a \omega^2\cos 2 \omega t$$

But for SHM,
$$a \propto - \omega^2$$

So, This is not $$SHM$$.
So, the correct option will be $$(D)$$
Question 9
1 / -0

A particle is vibrating in simple harmonic motion with amplitude of $$4\ cm$$. At what displacement from the equilibrium position is its energy half potential and half kinetic?

A
$$1\ cm$$

B
$$\sqrt {2}\ cm$$

C
$$2\ cm$$

D
$$2\sqrt {2}\ cm$$

Solution

$$Kinetic\,\,energy = \frac{1}{2}m{w^2}\left( {{A^2} - {x^2}} \right)$$
Where x is the displacement from mean position
$$\begin{array}{l} Potential\, \, energy=\frac { 1 }{ 2 } m{ w^{ 2 } }{ x^{ 2 } } \\ \frac { 1 }{ 2 } m{ w^{ 2 } }{ x^{ 2 } }=\frac { 1 }{ 2 } m{ w^{ 2 } }\left( { { A^{ 2 } }-{ x^{ 2 } } } \right) \\ { x^{ 2 } }={ A^{ 2 } }-{ x^{ 2 } } \\ { x^{ 2 } }-{ x^{ 2 } }.{ A^{ 2 } } \\ 2{ x^{ 2 } }={ 4^{ 2 } } \\ 2{ x^{ 2 } }=16 \\ { x^{ 2 } }=8 \\ x=2\sqrt { 2 } \, \, cm \end{array}$$
Option D.
Question 10
1 / -0

A particles starts its $$SHM$$ from mean position at $$t = 0$$. If its time period is $$T$$ and amplitude $$A$$. The distance travelled by the particle in the time from $$t = 0$$ to $$t = \dfrac{5T}{4} $$ is

A
$$A$$

B
$$2A$$

C
$$4A$$

D
$$5A$$

Solution