Waves Test

Result

Waves Test - 58

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Weekly Quiz Competition

Question 1
1 / -0

The equation of wave is given by $$y = 10 \sin(2\pi t/45 + \alpha)$$. If the displacement is$$ 5 cm$$ at $$=0$$, then the total phase at $$t = 7.5 s$$ will be:

A
$$\pi/3$$ rad

B
$$\pi/2$$ rad

C
$$2\pi/5$$ rad

D
$$2\pi/3$$ rad

Solution
Question 2
1 / -0

A $$4kg$$ particle is moving along the x-axis under the action of the force which varies with distance $$x$$ given by $$F=-\left( \dfrac { { \pi }^{ 2 } }{ 16 } \right) x$$ Newton. At $$t=2sec$$, the particle passes through the origin and at $$t=10sec$$, its speed is $$ 4\sqrt { 2 } m/s $$. The amplitude of the motion is

A
$$\dfrac { 32\sqrt { 2 } }{ \pi } m$$

B
$$\dfrac { 16 }{ \pi } m$$

C
$$\dfrac { 4 }{ \pi } m$$

D
$$\dfrac { 64\sqrt { 2 } }{ \pi } m $$
Question 3
1 / -0

What is the period of a vibrating particle executing S.H.M.. Which has acceleration $$ 0.12 m/s^2 $$ when its displacement is $$ 3 \times 10^{-2} m? $$

A
2 s

B
1 s

C
$$ 2 \pi s $$

D
$$ \pi s $$

Solution
Question 4
1 / -0

Two waves represented as
$$y_1 = a \sin (\omega t + {\pi}/6)$$, $$y_2 = a \cos \omega t$$
The resultant amplitude is-

A
$$a$$

B
$$a\sqrt { 2 } $$

C
$$a\sqrt {3}$$

D
$$2a$$

Solution
Question 5
1 / -0

Two particles in the path of a wave, of velocity $$360\ m/s$$ and frequency $$2000\ Hz$$, differ in phase by $$120^{o}$$. The distance between the particles is

A
$$6\ cm$$

B
$$12\ cm$$

C
$$18\ cm$$

D
$$100\ cm$$

Solution
Question 6
1 / -0

A particle executes simple harmonic motion and is located x = a, b and c at time$$ \ t_0 \ 2t_0 \ and \ 3t_0 $$ respectively.The frequency of the oscillation is:-

A
$$ \cfrac { 1 }{ 2\pi t_ 0 } cos^ {-1} \left( \frac { a+b }{ 2c } \right) $$

B
$$ \dfrac { 1 }{ 2\pi t_ 0 } cos^ {-1} \left( \frac { a+b }{ c } \right) $$

C
$$ \dfrac { 1 }{ 2\pi t_ 0 } cos^ {-1}\left( \frac { a+2b }{ 3c } \right) $$

D
$$ \dfrac { 1 }{ 2\pi t_ 0 } cos^{- 1}\left( \frac { a+c }{ 2b } \right) $$

Solution
Question 7
1 / -0

On the superposition of the two waves given as $${ y }_{ 1 }={ A }_{ 0 }sin\left( \omega t-kx \right) $$ and $${ y }_{ 2 }={ A }_{ 0 }cos\left( \omega t-kx+\dfrac { \pi }{ 6 } \right) $$. the resultant amplitude of oscillations will be

A
$$\sqrt { 3 } { A }_{ 0 }$$

B
$$\dfrac { { A }_{ 0 } }{ 2 } $$

C
$${ A }_{ 0 }$$

D
$$\dfrac { 3 }{ 2 } { A }_{ 0 }$$

Solution
Question 8
1 / -0

$$If x=a\sin { (\omega +\frac { \pi }{ 6} })$$ and $$x^2=a\cos \omega { t }$$ ,then what is the phase difference between the two waves:

A
$$\dfrac{\pi}{3}$$

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

C
$$\dfrac{\pi}{2}$$

D
$$\pi $$
Question 9
1 / -0

Two waves are represented by the equations
$$y_1 = a sin (\omega + kx + 0.57 )m$$ and
$$y_2 = a cos (\omega t + kx)m$$
where x is in meter and t in sec. The phase difference between them is

A
0.57 radian

B
1.0 radian

C
1,25 radian

D
1.57 radian

Solution

$$y_{1}=A sin (\omega t+kx+0.57)$$
$$y_{2}=A cos (\omega t+kx)$$
$$= A sin (\omega t+kx+\frac{\pi }{2})$$
phase difference = $$\frac{\pi }{2}-0.57$$
$$=1.57-0.57$$
= 1 radian
Option (B)
Question 10
1 / -0

Two particles undergo SHM along the same line with the same time period (T They will cross each other after a further time

A
$$t = \dfrac{{T}}{2}$$

B
$$t = \dfrac{T}{\sqrt {2}}$$

C
$$t= \dfrac{2T}{4}$$

D
$$t = \dfrac{3T}{8}$$

Solution

Hence, option $$(D)$$ is correct answer.