+91-8527521718
Signup | Login
  • 1(current)
  • 2
Q. No.

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%
Hints

A simple pendulum of mass \(m\) swings about point \(B\) between extreme positions \(A\) and \(C\). Net force acting on the bob at these three points is correctly shown by:

1. 2.
3. 4.
Subtopic:  Angular SHM |
Level 3: 35%-60%
Hints
Next Hint
If \(T_1,T_2,T_3,T_4\) and \(T_5\) represent the tension in the string of a simple pendulum when the bob is at the left extreme, right extreme, mean, any intermediate left and any intermediate right positions, respectively. Then, which of the following relations are correct?
(A) \(T_1=T_2\) (B) \(T_3>T_2\)
(C) \(T_4>T_3\) (D) \(T_3=T_4\)
(E) \(T_5>T_2\)
Choose the most appropriate answer from the options given below:
1. (A), (B) and (C) only 2. (B), (C) and (D) only
3. (A), (B) and (E) only 4. (C), (D) and (E) only
Subtopic:  Angular SHM |
 68%
Level 2: 60%+
NEET - 2024
Hints
advertisementadvertisement
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 |
 85%
Level 1: 80%+
Hints
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%+
Hints
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
advertisementadvertisement

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
Two simple pendulums of length \(1~\text{m}\) and \(16~\text{m}\) are in the same phase at the mean position at any instant. If \(T\) is the time period of the smaller pendulum, then the minimum time after which they will again be in the same phase will be:
1. \(\frac{3T}{2}\)
2. \(\frac{3T}{4}\)
3. \(\frac{2T}{3}\)
4. \(\frac{4T}{3}\)
Subtopic:  Angular SHM |
Level 3: 35%-60%
Hints

A simple pendulum attached to the ceiling of a stationary lift has a time period of 1 s. The distance y covered by the lift moving downward varies with time as y = 3.75 t2, where y is in meters and t is in seconds. If g = 10 m/s2, then the time period of the pendulum will be:

1. 4 s 2. 6 s
3. 2 s 4. 12 s
Subtopic:  Types of Motion | Simple Harmonic Motion | Angular SHM |
 62%
Level 2: 60%+
Hints
advertisementadvertisement

Two spherical bobs of masses \(M_A\) and \(M_B\) are hung vertically from two strings of length \(l_A\) and \(l_B\) respectively. If they are executing SHM with frequency as per the relation \(f_A=2f_B,\) Then:
1. \(l_A = \frac{l_B}{4}\)
2. \(l_A= 4l_B\)
3. \(l_A= 2l_B~\&~M_A=2M_B\)
4. \(l_A= \frac{l_B}{2}~\&~M_A=\frac{M_B}{2}\)

Subtopic:  Angular SHM |
 75%
Level 2: 60%+
AIPMT - 2000
Hints
  • 1(current)
  • 2
Q. No.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions

© 2025 GoodEd Technologies Pvt. Ltd.