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Oscillations MCQs for NEET — Physics Questions with Answers

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If the displacement equation of a particle be represented by $y = A \sin P T + B \cos P T$ , the particle executes

A uniform circular motion

A uniform elliptical motion

A S.H.M.

A rectilinear motion

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Explanation

$l e t A = r \cos θ, B = r \sin θ$

$\Rightarrow y = r \sin (P T + θ) w h i c h i s t h e e q u a t i o n o f S H M .$

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A particle with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force $Fsinωt$ . If the amplitude of the particle is maximum for $ω = ω_{1}$ and the energy of the particle is maximum for $ω = ω_{2}$ , then (where $ω_{0}$ is natural frequency of oscillation of particle)

$ω_{1} = ω_{0}$ and $ω_{2} \neq ω_{0}$

$ω_{1} = ω_{0}$ and $ω_{2} = ω_{0}$

$ω_{1} \neq ω_{0}$ and $ω_{2} = ω_{0}$

$ω_{1} \neq ω_{0}$ and $ω_{2} \neq ω_{0}$

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Explanation

Energy of particle is maximum at natural frequency i.e., $ω_{2} = ω_{0}$ .

For amplitude resonance (amplitude maximum)

$A = \frac{F_{o}}{\sqrt{m^{2} {(ω_{o}^{2} - ω_{d}^{2})}^{2} + {(b ω_{d})}^{2}}} F o r m a x i m u m A, \frac{d A}{d ω} = 0 S o l v i n g, ω_{o}^{2} - ω_{d}^{2} = \frac{b^{2}}{2 m^{2}} S o, ω_{1} < ω_{o} o r ω_{1} \neq ω_{o}$

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The displacement of a particle varies according to the relation $x = 4 (cosπ t + sinπ t) .$ The amplitude of the particle is

– 4

$4 \sqrt{2}$

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Explanation

(d) For given relation

Resultant amplitude= $\sqrt{4^{2} + 4^{2}}$ = $4 \sqrt{2}$

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A S.H.M. is represented by $x = 5 \sqrt{2} (\sin 2 πt + \cos 2 πt) .$ The amplitude of the S.H.M. is

10 cm

20 cm

$5 \sqrt{2}$ cm

50 cm

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Explanation

(a) $x = 5 \sqrt{2} (\sin 2 πt + \cos 2 πt) .$

$= 5 \sqrt{2} \sin 2 π t + 5 \sqrt{2} \cos 2 π t$

$x = 5 \sqrt{2} \sin 2 πt + 5 \sqrt{2} \sin (2 π t + \frac{π}{2})$

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The displacement of a particle varies with time as $x = 12 \sin w t - 16 \sin^{3} w t$ (in cm). If its motion is S.H.M., then its maximum acceleration is -

$12 ω^{2}$

$36 ω^{2}$

$144 ω^{2}$

$\sqrt{192} ω^{2}$

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Explanation

(b) $x = 12 \sin ω t - 16 \sin^{3} ω t$ = $4 [3 \sin ω t - 4 \sin^{3} ω t]$

= $4 [\sin 3 ω t] (b y u \sin g \sin 3 θ = 3 \sin θ - 4 \sin^{3} θ)$

$A c c e l e r a t i o n i s m a x i m u m w h e n x = A S o m a x i m u m a c c e l e r a t i o n : a_{m a x} = {(3 ω)}^{2} \times 4 = 36 ω^{2}$

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A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is $U (x) = k {[x]}^{3}$ , where k is a positive constant. If the amplitude of oscillation is a, then its time period T is -

Proportional to $\frac{1}{\sqrt{a}}$

Independent of a

C.
Proportional to $\sqrt{a}$
D.

Proportional to $a^{3 / 2}$

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Explanation

(a)

$U = k {|x|}^{3} \Rightarrow F = - \frac{d U}{d x} = - 3 k {|x|}^{2} A c c e l e r a t i o n; a = - ω^{2} x \Rightarrow F = m a = \frac{d^{2} x}{d t^{2}} = - m ω^{2} x O n c o m p a r i n g : m ω^{2} = 3 k x \Rightarrow ω = \sqrt{\frac{3 k x}{m}} \Rightarrow T = \frac{2 π}{ω} = 2 π \sqrt{\frac{m}{3 kx}} Also, for SHM, x = asinωt \Rightarrow T = 2 π \sqrt{\frac{m}{3 k a s \sin ω t}}$

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The metallic bob of a simple pendulum has the relative density $ρ$ . The time period of this pendulum is T. If the metallic bob is immersed in water, then the new time period is given by

$T \frac{ρ - 1}{ρ}$

$T \frac{ρ}{ρ - 1}$

$T \sqrt{\frac{ρ - 1}{ρ}}$

$T \sqrt{\frac{ρ}{ρ - 1}}$

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Explanation

(d) When the bob is immersed in water ,

its effective weight =

$(m g - \frac{m}{ρ} g) = m g (\frac{ρ - 1}{ρ}) s o g_{e f f} = g (\frac{ρ - 1}{ρ})$

$\frac{T}{T} = \sqrt{\frac{g}{g_{e f f}}} \Rightarrow T' = T \sqrt{\frac{ρ}{ρ - 1}}$

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The period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $θ$ , is given by -

$2 π \sqrt{\frac{L}{gcosθ}}$

$2 π \sqrt{\frac{L}{gsin θ}}$

$2 π \sqrt{\frac{L}{g}}$

$2 π \sqrt{\frac{L}{gtanθ}}$

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Explanation

When a simple pendulum is suspended from the roof of a vehicle moving down an inclined plane without friction, the effective acceleration due to gravity experienced by the pendulum is g cos(Î¸), where Î¸ is the angle of inclination. Therefore, the time period is T = 2Ï€âˆš(L/(g cos(Î¸))), where L is the length of the pendulum.

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One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to massless spring of spring constant K. A mass m hangs freely from the free end of the spring. The area of cross-section and Young's modulus of the wire is A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to -

$2 π (\frac{m}{K})$

$2 π {\{\frac{(YA + KL) m}{YAK}\}}^{1 / 2}$

$2 π \frac{mYA}{KL}$

$2 π \frac{mL}{YA}$

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Explanation

b) The wire may be treated as a string for which force constant $k_{1} = \frac{F o r c e}{E x t e n c i o n} = \frac{Y A}{L} (Y = \frac{F}{A} \times \frac{L}{∆ L})$

Spring constant of the spring $k_{2} = K$

Hence spring constant of the combination (series)

$k_{e q} = \frac{k_{1} k_{2}}{k_{1} - k_{2}} = (\frac{(Y A / L) K}{(Y A / L) + K}) = \frac{Y A K}{Y A + K L}$

$⊝$ Time period $T = 2 π \sqrt{\frac{m}{k}} = 2 π {[\frac{(YA + KL) m}{YAK}]}^{1 / 2}$

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A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency $ω_{0}$ . An external force F (t) proportional to $\cos ω t (ω \neq ω_{0})$ is applied to the oscillator. The time displacement of the oscillator will be proportional to -

$\frac{m}{{ω^{2}}_{0} - ω^{2}}$

$\frac{1}{m ({ω^{2}}_{0} - ω^{2})}$

$\frac{1}{m ({ω^{2}}_{1} + ω^{2})}$

$\frac{m}{{ω^{2}}_{1} + ω^{2}}$

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Explanation

(b) For forced oscillation,

$x = x_{0} \sin (ω t + \emptyset) a n d F = F_{0} \cos ω t$

where, $x_{0} = \frac{F}{m ({ω^{2}}_{0} - ω^{2})} \propto \frac{1}{m ({ω^{2}}_{0} - ω^{2})}$

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Frequently Asked Questions

Yes. You can attempt every Oscillations question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.