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The uniform stick of mass m length \(\text L\) is pivoted at the centre. In the equilibrium position shown in the figure, the identical light springs have their natural length. If the stick is turned through a small angle θ, it executes SHM. The frequency of the motion is:

 

1. \(\dfrac{1}{2 \pi} \sqrt{\dfrac{6 K}{m}} \)

2. \(\dfrac{1}{2 \pi} \sqrt{\dfrac{3 K}{2 m}} \)

3. \(\dfrac{1}{2 \pi} \sqrt{\dfrac{3 K}{m}} \)

4. None of these

Subtopic:  Angular SHM |
Level 3: 35%-60%
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A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector \(\vec a\) is correctly shown in: 

1. 2.
3. 4.
Subtopic:  Angular SHM |
Level 3: 35%-60%
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There is a simple pendulum hanging from the ceiling of a lift. When the lift is stand still, the time period of the pendulum is T. If the resultant acceleration becomes g/4, then the new time period of the pendulum is 

1. 0.8 T

2. 0.25 T

3. 2 T

4. 4 T

Subtopic:  Angular SHM |
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Level 1: 80%+
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­­A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. If the lift accelerates upwards with an acceleration g4 , then the period of the pendulum will be

1. T

2. T4

3. 2T5

4. 2T5

Subtopic:  Angular SHM |
 84%
Level 1: 80%+
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The bob of a pendulum of length l is pulled aside from its equilibrium position through an angle θ and then released. The bob will then pass through its equilibrium position with a speed v, where v equals

1. 2gl(1-sinθ)

2. 2gl(1+cosθ)

3. 2gl(1-cosθ)

4. 2gl(1+sinθ)

Subtopic:  Angular SHM |
 78%
Level 2: 60%+
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In a simple pendulum, the period of oscillation \(T\) is related to the length of the pendulum \(L\) as:
1. \(\frac{L}{T}= \text{constant}\)
2. \(\frac{L^2}{T}= \text{constant}\)
3. \(\frac{L}{T^2}= \text{constant}\)
4. \(\frac{L^2}{T^2}= \text{constant}\)
Subtopic:  Angular SHM |
 86%
Level 1: 80%+
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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 |
 84%
Level 1: 80%+
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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 |
 64%
Level 2: 60%+
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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 |
 84%
Level 1: 80%+
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A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is \(20\text{ m/s}^2\) at a distance of \(5\text{ m}\) from the mean position. The time period of oscillation is:
1. \(2\pi \text{ s}\)
2. \(\pi \text{ s}\)
3. \(2 \text{ s}\)
4. \(1 \text{ s}\)

Subtopic:  Angular SHM |
 86%
Level 1: 80%+
NEET - 2018
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