Blog Log in Get Started Free

Oscillations MCQs for NEET — Physics Questions with Answers

Q: Are these Oscillations questions free to attempt?

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.

Q: Do I need an account to practise Oscillations MCQs?

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

Q: Are these previous year NEET questions?

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

Practice free Oscillations (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

Chapter Difficulty Show only questions I answered incorrectly Clear Register free for difficulty & keyword filters

● ● ● ● ●

A body is executing Simple Harmonic Motion. At a displacement x its potential energy is $E_{1}$ and at a displacement y its potential energy is $E_{2}$ . The potential energy E at displacement $(x + y)$ is

$E = \sqrt{E_{1}} + \sqrt{E_{2}}$

$\sqrt{E} = \sqrt{E_{1}} + \sqrt{E_{2}}$

$E = E_{1} + E_{2}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b) $E_{1} = \frac{1}{2} K x^{2} \Rightarrow x = \sqrt{\frac{2 E_{1}}{K}}, E_{2} = \frac{1}{2} K y^{2} \Rightarrow y = \sqrt{\frac{2 E_{2}}{K}}$

$a n d E = \frac{1}{2} K {(x + y)}^{2} \Rightarrow x + y = \sqrt{\frac{2 E}{K}}$

$\Rightarrow \sqrt{\frac{2 E_{1}}{K}} + \sqrt{\frac{2 E_{2}}{K}} = \sqrt{\frac{2 E}{K}} \Rightarrow \sqrt{E_{1}} + \sqrt{E_{2}} = \sqrt{E}$

● ● ● ● ●

The equation of motion of a particle is $\frac{d^{2} y}{d t^{2}} + K y = 0$ where K is positive constant. The time period of the motion is given by

$\frac{2 π}{K}$

$2 πK$

$\frac{2 π}{\sqrt{K}}$

$2 π \sqrt{K}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(c) On comparing with standard equation $\frac{d^{2} y}{d t^{2}} + ω^{2} y = 0$ we get $ω^{2} = K \Rightarrow ω \frac{2 π}{T} = \sqrt{K} \Rightarrow T = \frac{2 π}{\sqrt{K}}$

● ● ● ● ●

The kinetic energy of a particle executing S.H.M. is 16 J when it is in its mean position. If the amplitude of oscillations is 25 cm and the mass of the particle is 5.12 kg, the time period of its oscillation is -

$\frac{π}{5} s e c$

$2 π \sec$

$20 π \sec$

$5 π \sec$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

a) At mean position, the kinetic energy is maximum.

Hence $\frac{1}{2} m a^{2} ω^{2} = 16$

On putting the values we get $ω = 10 \Rightarrow T = \frac{2 π}{w} = \frac{π}{5} s e c$

● ● ● ● ●

A pendulum has time period T. If it is taken on to another planet having acceleration due to gravity half and mass 9 times that of the earth then its time period on the other planet will be

$\sqrt{T}$

$T^{1 / 3}$

$\sqrt{2}$ T

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d) $T = 2 π \sqrt{\frac{l}{g}} \Rightarrow T \propto \frac{1}{\sqrt{g}} \Rightarrow \frac{T_{p}}{T_{e}} = \sqrt{\frac{g_{e}}{g_{p}}} = \sqrt{\frac{2}{1}} \Rightarrow T_{p} = \sqrt{2} T$

● ● ● ● ●

A particle in SHM is described by the displacement equation $x (t) = A \cos (ω t + θ) . I f t h e i n i t i a l$ position of the particle is 1 cm and its initial velocity is $π$ cm/s, what is its amplitude? (The angular frequency of the particle is $π s^{- 1}$ )

1 cm

$\sqrt{2}$ cm

2 cm

2.5 cm

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b) Given, $v = πcm / \sec, x = 1 cm and ω = {πs}^{2} using v = ω \sqrt{a^{2} - x^{2}} \Rightarrow π = π \sqrt{a^{2} - 1} \Rightarrow 1 = a^{2} - 1 \Rightarrow a = \sqrt{2} cm .$

● ● ● ● ●

A simple pendulum hanging from the ceiling of a stationary lift has a time period T₁. When the lift moves downward with constant velocity, the time period is T₂, then

$T_{2}$ is infinity

$T_{2} > T_{1}$

$T_{2} < T_{1}$

$T_{2} = T_{1}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d) $T \propto \frac{1}{\sqrt{g}}$ and g is same in both cases so time period remains same.

● ● ● ● ●

If the length of a pendulum is made 9 times and mass of the bob is made 4 times , then the value of time period becomes

3/2T

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(a) $T = 2 π \sqrt{\frac{l}{g}} \Rightarrow T \propto \sqrt{l}$ , hence if l made 9 times, T becomes 3 times.

Also time period of simple pendulum does not depends on the mass of the bob.

● ● ● ● ●

The period of a simple pendulum measured inside a stationary lift is found to be T. If the lift starts accelerating upwards with acceleration of g/3 then the time period of the pendulum is

$\frac{T}{\sqrt{3}}$

$\frac{T}{3}$

$\frac{\sqrt{3}}{2} T$

$\sqrt{3} T$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

For ascending lift with acceleration a,

$T_{2} = 2 π \sqrt{\frac{l}{g + a}} \Rightarrow \frac{T_{1}}{T_{2}} = \sqrt{\frac{g + a}{g}} \Rightarrow \frac{T}{T_{2}} = \sqrt{\frac{g + \frac{g}{3}}{g}} = \sqrt{\frac{4}{3}} \Rightarrow T_{2} = \frac{\sqrt{3}}{2} T$

● ● ● ● ●

The time period of a simple pendulum of length L as measured in an elevator descending with acceleration $\frac{g}{3}$ is

$2 π \sqrt{\frac{3 L}{g}}$

$π \sqrt{(\frac{3 L}{g})}$

$2 π \sqrt{(\frac{3 L}{2 g})}$

$2 π \sqrt{\frac{2 L}{3 g}}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(c) The effective acceleration in a lift descending with acceleration $\frac{g}{3} is g_{eff} = g - \frac{g}{3} = \frac{2 g}{3} so, T = 2 π \sqrt{(\frac{L}{g_{eff}})} = 2 π \sqrt{(\frac{L}{2 g / 3})} = 2 π \sqrt{(\frac{3 L}{2 g})}$

● ● ● ● ●

If a body is released into a tunnel dug across the diameter of earth, it executes simple harmonic motion with time period

$T = 2 π \sqrt{\frac{R_{e}}{g}}$

$T = 2 π \sqrt{\frac{2 R_{e}}{g}}$

$T = 2 π \sqrt{\frac{R_{e}}{2 g}}$

T=2 seconds

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(a)

$A c c e l e r a t i o n d u e t o g r a v i t y a t a d e p t h d b e l o w t h e s u r f a c e o f t h e e a r t h; g' = g (1 - \frac{d}{R}) = \frac{g}{R} (R - d) = \frac{g}{R} x g' \propto x \Rightarrow ω^{2} = \frac{g}{R} T = 2 π \sqrt{\frac{R_{e}}{g}}$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

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.