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Physics NEET Questions — Free MCQ Practice

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A point performs simple harmonic oscillation of period T and the equation of motion is given by x= a sin $(ω t + π / 6)$ .After the elapse of what fraction of the time period the velocity of the point will be equal to half to its maximum velocity?

A.

$\frac{T}{8}$

B.

$\frac{T}{6}$

C.

$\frac{T}{3}$

D.

$\frac{T}{12}$

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Explanation

Velocity is the time derivation of displacement. Writing the given equation of a point performing SHM

x= a sin $(ω t + \frac{π}{6})$ ..(i)

Differentiating Eq. (i),w.r.t. time, we obtain

v= $\frac{d x}{d t} = a ω \cos (ω t + \frac{π}{6})$

It is given that v= $\frac{a ω}{2}$ ,so that

$\frac{a ω}{2} = a ω \cos (ω t + \frac{π}{6}) o r \frac{1}{2} = \cos (ω t + \frac{π}{6}) o r \cos \frac{π}{3} = \cos (ω t + \frac{π}{6}) o r ω t + \frac{π}{6} = \frac{π}{3} \Rightarrow ω t = \frac{π}{6} o r t = \frac{π}{6 ω} = \frac{π \times T}{6 \times 2 π} = \frac{T}{12}$

Thus, at $\frac{T}{12}$ velocity of the point will be equal to half of its maximum velocity.

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A S.H.M. has amplitude ‘a’ and time period T. The maximum velocity will be -

A.

$\frac{4 a}{T}$

B.

$\frac{2 a}{T}$

C.

$2 π \sqrt{\frac{a}{T}}$

D.

$\frac{2 πa}{T}$

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Explanation

(d) $v_{m a x} = a w = \frac{a . 2 π}{T} = \frac{2 πa}{T}$

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Two particles P and Q start from origin and execute Simple Harmonic Motion along X-axis with same amplitude but with periods 3 seconds and 6 seconds respectively. The ratio of the velocities of P and Q when they meet is -

A.

1 : 2

B.

2 : 1

C.

2 : 3

D.

3 : 2

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Explanation

(b) The particles will meet at the mean position when P completes one oscillation and Q completes half an oscillation

$s o \frac{v_{p}}{v_{q}} = \frac{a w_{p}}{a w_{q}} = \frac{T q}{T_{p}} = \frac{6}{3} = \frac{2}{1}$

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The amplitude of a particle executing SHM is 4 cm. At the mean position the speed of the particle is 16 cm/sec. The distance of the particle from the mean position at which the speed of the particle becomes $8 \sqrt{3} c m / s e c$ will be

A.

$2 \sqrt{3} c m$

B.

$\sqrt{3} c m$

C.

1 cm

D.

2 cm

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Explanation

(d) At mean position velocity is maximum

i.e., $v_{m a x} = w a \Rightarrow w \frac{v_{m a x}}{a} = \frac{16}{4} = 4$

$v = w \sqrt{a^{2} - y^{2}} \Rightarrow 8 \sqrt{3} = 4 \sqrt{4^{2} - y^{2}}$

$\Rightarrow 192 = 16 (16 - y^{2}) \Rightarrow 12 = 16 - y^{2} \Rightarrow y = 2 c m$

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The maximum velocity of a simple harmonic motion represented by $y = 3 \sin (100 t + \frac{π}{6})$ is given by

A.

300

B.

$\frac{3 π}{6}$

C.

100

D.

$\frac{π}{6}$

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Explanation

(a) $v_{m a x} = a w = 3 \times 100 = 300$

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The instantaneous displacement of a simple pendulum oscillator is given by $x = A \cos (ω t + \frac{π}{4})$ . Its speed will be maximum at time

A.

$\frac{π}{4 ω}$

B.

$\frac{π}{2 ω}$

C.

$\frac{π}{ω}$

D.

$\frac{2 π}{ω}$

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Explanation

(a) $x = A \cos (ω t + \frac{π}{4}) a n d v = \frac{d x}{d t} = - a w \sin (ω t + \frac{π}{4})$

For maximum speed,

$\sin (ω t + \frac{π}{4}) = 1 \Rightarrow ω t + \frac{π}{4} = \frac{π}{2} o r ω t = \frac{π}{2} - \frac{π}{4} \Rightarrow t = \frac{π}{4 ω}$

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The displacement of a particle moving in S.H.M. at any instant is given by $y = a \sin$ $ω t$ . The acceleration after time $t = \frac{T}{4}$ (where T is the time period) -

A.

$a ω$

B.

$- a ω$

C.

$a ω^{2}$

D.

$- a ω^{2}$

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Explanation

At t= T/4, particle is at y=a

Acceleration = $- a ω^{2}$ when it is at positive extreme point.

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The displacement of an oscillating particle varies with time (in seconds) according to the equation $y (c m) = \sin \frac{π}{2} (\frac{t}{2} + \frac{1}{3})$ .The maximum acceleration of the particle is approximately

A.

$5.21 c m / s^{2}$

B.

$3.62 c m / s^{2}$

C.

$1.81 c m / s^{2}$

D.

$0.62 c m / s^{2}$

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Explanation

(d) $a_{m a x} = ω^{2} A = {(\frac{π}{4})}^{2} \times 1 = 0.62 c m / s e c^{2}$

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A particle moving along the x-axis executes simple harmonic motion, then the force acting on it is given by

A.

– A Kx

B.

A cos (Kx)

C.

A exp (– Kx)

D.

A Kx

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Explanation

(a) For S.H.M. $F = - k x$

so Force = Mass $\times$ Acceleration $\propto - x$

$\Rightarrow$ F = – Akx; where A and k are positive constants

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What is the maximum acceleration of the particle doing the SHM $y = 2 \sin [\frac{πt}{2} + ϕ]$ where 2 is in cm

A.

$\frac{π}{2} c m / s^{2}$

B.

$\frac{π^{2}}{2} c m / s^{2}$

C.

$\frac{π}{4} c m / s^{2}$

D.

$\frac{π}{4} c m / s^{2}$

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Explanation

(b) Comparing given equation with standard equation

$y = a \sin (ω t + ϕ), w e g e t, a = 2 c m ω = \frac{π}{2} s o A_{m a x} = ω^{2} A = {(\frac{π}{2})}^{2} \times 2 = \frac{π^{2}}{2} c m / s^{2}$

Physics MCQs for NEET — Practice Questions with Answers

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