Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly

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:
1. \(\sqrt{T} \) 2. \(T \)
3. \({T}^{1 / 3} \) 4. \(\sqrt{2} {T}\)

Subtopic:  Angular SHM |
 83%
Level 1: 80%+
Hints

Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly

A particle in SHM is described by the displacement equation xt=Acosωt+θ. If the initial position of the particle is 1 cm and its initial velocity is πcm/s, what is its amplitude? (The angular frequency of the particle is πs-1)

1. 1 cm       
2. 2 cm
3.  2 cm     
4. 2.5 cm

Subtopic:  Simple Harmonic Motion |
 58%
Level 3: 35%-60%
Hints

Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly
A simple pendulum hanging from the ceiling of a stationary lift has a time period \(T_1\). When the lift moves downward with constant velocity, then the time period becomes \(T_2\). It can be concluded that: 
1. \(T_2 ~\text{is infinity} \) 2. \(T_2>T_1 \)
3. \(T_2<T_1 \) 4. \(T_2=T_1\)
Subtopic:  Angular SHM |
 63%
Level 2: 60%+
Hints

Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly

If the length of a pendulum is made \(9\) times and the mass of the bob is made \(4\) times, then the value of time period will become:
1. \(3T\)
2. \(\dfrac{3}{2}{T}\)
3. \(4{T}\)
4. \(2{T}\)

Subtopic:  Angular SHM |
 83%
Level 1: 80%+
Hints

Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly

A simple harmonic wave having an amplitude a and time period T is represented by the equation y=5sinπt+4m Then the value of amplitude (a) in (m) and time period  (T) in second are       

1.   a=10,T=2   

2. a=5,T=1

3.    a=10,T=1    

4. a=5,T=2

Subtopic:  Simple Harmonic Motion |
 85%
Level 1: 80%+
Hints

Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly

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

1. T3

2. T3

3. 32T

4. 3T

Subtopic:  Simple Harmonic Motion |
 87%
Level 1: 80%+
Hints

Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly

The time period of a simple pendulum of length L as measured in an elevator descending with acceleration g3 is

1. 2π3Lg

2. π3Lg

3. 2π3L2g

4. 2π2L3g

Subtopic:  Simple Harmonic Motion |
 83%
Level 1: 80%+
Hints

Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly
If a body is released into a tunnel dug across the diameter of earth, it executes simple harmonic motion with time period:
1. \(T = 2\pi \sqrt{\frac{R_e}{g}}\)
2. \(T = 2\pi \sqrt{\frac{2R_e}{g}}\)
3. \(T = 2\pi \sqrt{\frac{R_e}{2g}}\)
4. \(T = 2~\text{s}\)
Subtopic:  Simple Harmonic Motion |
 69%
Level 2: 60%+
Hints

Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly

If the displacement equation of a particle be represented by y=AsinPt+BcosPt , the particle executes

1.         A uniform circular motion

2.         A uniform elliptical motion

3.         A S.H.M.

4         A rectilinear motion

Subtopic:  Simple Harmonic Motion |
 82%
Level 1: 80%+
Hints

Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly

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=ω1  and the energy of the particle is maximum for ω=ω2, then (where ω0 is natural frequency of oscillation of particle)

1. ω1=ω0 and ω2ω0

2. ω1=ω0 and ω2=ω0

3. ω1ω0 and ω2=ω0

4. ω1ω0 and ω2ω0

Level 3: 35%-60%
Hints