Two pendulums suspended from the same point have lengths of \(2\) m and \(0.5\) m. If they are displaced slightly and released, then they will be in the same phase when the small pendulum has completed:
1. \(2\) oscillations
2. \(4\) oscillations
3. \(3\) oscillations
4. \(5\) oscillations

Subtopic:  Simple Harmonic Motion |
 66%
From NCERT
AIPMT - 1998
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If the time of mean position from amplitude (extreme) position is \(6\) seconds, then the frequency of SHM will be:
1. \(0.01~\text{Hz}\) 2. \(0.02~\text{Hz}\)
3. \(0.03~\text{Hz}\) 4. \(0.04~\text{Hz}\)
Subtopic:  Simple Harmonic Motion |
 68%
From NCERT
AIPMT - 1998
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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 |
 81%
From NCERT
AIPMT - 1998
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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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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 |
 72%
From NCERT
AIPMT - 2000
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The total energy of the particle performing SHM depends on: 
1. \(k,\) \(a,\) \(m\)
2. \(k,\) \(a\)
3. \(k,\) \(a\)\(x \)
4. \(k,\) \(x \)

Subtopic:  Energy of SHM |
 68%
From NCERT
AIPMT - 2001
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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. t02=t12+t22

2. t0-2=t1-2+t2-2

3. t0-1=t1-1+t2-1

4. t0=t1+t2

Subtopic:  Combination of Springs |
 69%
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. \(\pm \frac{a}{2}\)
2. \(+a\)
3. \(\pm a\)
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. \(\frac{T}{4}\)
2. \(T\)
3. \(\frac{T}{2}\)
4. \(2T\)

Subtopic:  Spring mass system |
 73%
From NCERT
AIPMT - 2003
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