# 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 circular motion of a particle with constant speed is:

 1 Periodic and simple harmonic 2 Simple harmonic but not periodic 3 Neither periodic nor simple harmonic 4 Periodic but not simple harmonic
Subtopic:  Types of Motion |
79%
From NCERT
AIPMT - 2005
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The frequency of a spring is $$n$$ after suspending mass $$M.$$ Now, after mass $$4M$$ mass is suspended from the spring, the frequency will be:

 1 $$2n$$ 2 $$n/2$$ 3 $$n$$ 4 none of the above

Subtopic:  Spring mass system |
80%
From NCERT
AIPMT - 1998
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Which one of the following statements is true for the speed 'v' and the acceleration 'a' of a particle executing simple harmonic motion?

 1 The value of a is zero whatever may be the value of 'v'. 2 When 'v' is zero, a is zero. 3 When 'v' is maximum, a is zero. 4 When 'v' is maximum, a is maximum.
Subtopic:  Simple Harmonic Motion |
86%
From NCERT
AIPMT - 2004
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A spring elongates by a length 'L' when a mass 'M' is suspended to it. Now a tiny mass 'm' is attached to the mass 'M' and then released. The new time period of oscillation will be:

1.  $$2 \pi \sqrt{\frac{\left(\right. M + m \left.\right) l}{Mg}}$$

2. $$2 \pi \sqrt{\frac{ml}{Mg}}$$

3. $$2 \pi \sqrt{L / g}$$

4. $$2 \pi \sqrt{\frac{Ml}{\left(\right. m + M \left.\right) g}}$$

Subtopic:  Spring mass system |
59%
From NCERT
AIPMT - 1999
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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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When a mass is suspended separately by two different springs, in successive order, then the time period of oscillations is $$t _1$$ and $$t_2$$ respectively. If it is connected by both springs as shown in the figure below, then the time period of oscillation becomes $$t_0.$$ The correct relation between $$t_0,$$ $$t_1$$ & $$t_2$$ is:

1. ${{\mathrm{t}}_{0}}^{2}={{\mathrm{t}}_{1}}^{2}+{{\mathrm{t}}_{2}}^{2}$

2. ${{\mathrm{t}}_{0}}^{-2}={{\mathrm{t}}_{1}}^{-2}+{{\mathrm{t}}_{2}}^{-2}$

3. ${{\mathrm{t}}_{0}}^{-1}={{\mathrm{t}}_{1}}^{-1}+{{\mathrm{t}}_{2}}^{-1}$

4. ${\mathrm{t}}_{0}={\mathrm{t}}_{1}+{\mathrm{t}}_{2}$

Subtopic:  Combination of Springs |
68%
From NCERT
AIPMT - 2002
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The displacement between the maximum potential energy position and maximum kinetic energy position for a particle executing simple harmonic motion is:

1. $±\frac{\mathrm{a}}{2}$

2. $+\mathrm{a}$

3. $±\mathrm{a}$

4. $-1$

Subtopic:  Energy of SHM |
73%
From NCERT
AIPMT - 2002
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The time period of a mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be:
1. T/4
2. T
3. T/2
4. 2T

Subtopic:  Spring mass system |
72%
From NCERT
AIPMT - 2003
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A particle of mass m oscillates with simple harmonic motion between points x1 and x2, the equilibrium position being O. Its potential energy is plotted. It will be as given below in the graph:

 1 2 3 4
Subtopic:  Energy of SHM |
84%
From NCERT
AIPMT - 2003
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