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 will become:

1. 3T

2. 3/2T

3. 4T

4. 2T

Subtopic:  Angular SHM |
82%
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In a simple pendulum, the period of oscillation $$T$$ is related to 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 |
84%
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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 $$\sqrt{\mathrm{T}}$$ 2 $$T$$ 3 $$\mathrm{T}^{1 / 3}$$ 4 $$\sqrt{2} \mathrm{~T}$$
Subtopic:  Angular SHM |
82%
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A spring pendulum is on the rotating table. The initial angular velocity of the table is $$\omega_{0}$$ and the time period of the pendulum is $$T_{0}.$$ Now the angular velocity of the table becomes $$2\omega_{0},$$ then the new time period will be:
1.
$$2T_{0}$$
2. $$T_0\sqrt{2}$$
3. remains the same
4. $$\frac{T_0}{\sqrt{2}}$$

Subtopic:  Angular SHM |
70%
From NCERT
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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. ${l}_{A}=4{l}_{B}$

3. ${l}_{A}=2{l}_{B}$ $&$ ${M}_{A}=2{M}_{B}$

4. ${l}_{A}=\frac{{l}_{B}}{2}$ $&$ ${M}_{A}=\frac{{M}_{B}}{2}$

Subtopic:  Angular SHM |
72%
From NCERT
AIPMT - 2000
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The frequency of a simple pendulum in a free-falling lift will be:
1. zero
2. infinite
3. can't say
4. finite

Subtopic:  Angular SHM |
67%
From NCERT
AIPMT - 1999
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A simple pendulum is pushed slightly from its equilibrium towards left and then set free to execute simple harmonic motion. Select the correct graph between its velocity($$v$$) and displacement ($$x$$).

 1 2 3 4
Subtopic:  Angular SHM |
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The period of oscillation of a simple pendulum of length $$\mathrm{L}$$ suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $\theta$, is given by:

1.   $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{gcos\theta }}}$

2.  $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{gsin}\theta }}$

3.

4. $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{gtan\theta }}}$

Subtopic:  Angular SHM |
60%
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A simple pendulum hanging from the ceiling of a stationary lift has a time period T1. When the lift moves downward with constant velocity, then the time period becomes T2. It can be concluded that:

 1 $$T_2 ~\text{is infinity}$$ 2 $$\mathrm{T}_2>\mathrm{T}_1$$ 3 $$\mathrm{T}_2<\mathrm{T}_1$$ 4 $$T_2=T_1$$
Subtopic:  Angular SHM |
61%
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Two simple pendulums of length 1 m and 16 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.  $\frac{3\mathrm{T}}{2}$

2.  $\frac{3\mathrm{T}}{4}$

3.  $\frac{2\mathrm{T}}{3}$

4.  $\frac{4\mathrm{T}}{3}$

Subtopic:  Angular SHM |
From NCERT
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