The amplitude and the time period in an SHM are \(0.5\) cm and \(0.4\) sec respectively. If the initial phase is \(\frac{\pi}{2}\) radian, then the equation of SHM will be:
1. \(y = 0.5\sin(5\pi t)\)
2. \(y = 0.5\sin(4\pi t)\)
3. \(y = 0.5\sin(2.5\pi t)\)
4. \(y = 0.5\cos(5\pi t)\)

Subtopic:  Linear SHM |
 71%
Level 2: 60%+
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A mass m is suspended from two springs of spring constant k1 and k2 as shown in the figure below. The time period of vertical oscillations of the mass will be
                

1. 2πk1+k2m

2. 2πmk1+k2

3. 2πm(k1k2)k1+k2

4. 2πmk1+k2k1k2

Subtopic:  Combination of Springs |
 80%
Level 1: 80%+
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A block of mass \(4~\text{kg}\) hangs from a spring of spring constant \(k = 400~\text{N/m}\). The block is pulled down through \(15~\text{cm}\) below the equilibrium position and released. What is its kinetic energy when the block is \(10~\text{cm}\) below the equilibrium position? [Ignore gravity]
1. \(5~\text{J}\)
2. \(2.5~\text{J}\)
3. \(1~\text{J}\)
4. \(1.9~\text{J}\)

Subtopic:  Energy of SHM |
 79%
Level 2: 60%+
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The radius of the circle, the period of revolution, initial position and direction of revolution are indicated in the figure.

The \(y\)-projection of the radius vector of rotating particle \(P\) will be:

1. \(y(t)=3 \cos \left(\dfrac{\pi \mathrm{t}}{2}\right)\), where \(y\) in m
2. \(y(t)=-3 \cos 2 \pi t\) , where \(y\) in m
3. \(y(t)=4 \sin \left(\dfrac{\pi t}{2}\right)\), where \(y\) in m
4. \(y(t)=3 \cos \left(\dfrac{3 \pi \mathrm{t}}{2}\right) \),  where \(y\) in m
Subtopic:  Phasor Diagram |
 78%
Level 2: 60%+
NEET - 2019
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A particle executing simple harmonic motion has a kinetic energy of \(K_0 \cos^2(\omega t)\). The values of the maximum potential energy and the total energy are, respectively:
1. \(0~\text{and}~2K_0\)
2. \(\frac{K_0}{2}~\text{and}~K_0\)
3. \(K_0~\text{and}~2K_0\)
4. \(K_0~\text{and}~K_0\)
Subtopic:  Energy of SHM |
 64%
Level 2: 60%+
AIPMT - 2007
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A point performs simple harmonic oscillation of period \(\mathrm{T}\) and the equation of motion is given by; \(x=a \sin (\omega t+\pi / 6)\)After the elapse of what fraction of the time period, the velocity of the point will be equal to half of its maximum velocity?
1. \( \frac{T}{8} \)

2. \( \frac{T}{6} \)

3. \(\frac{T}{3} \)

4. \( \frac{T}{12}\)

Subtopic:  Linear SHM |
 71%
Level 2: 60%+
AIPMT - 2008
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The angular velocities of three bodies in simple harmonic motion are \(\omega_1, \omega_2, \omega_3\) with their respective amplitudes as \(A_1, A_2, A_3.\) If all the three bodies have the same mass and maximum velocity, then:
1. \(A_1 \omega_1=A_2 \omega_2=A_3 \omega_3\)
2. \(A_1 \omega_1^2=A_2 \omega_2^2=A_3 \omega_3^2\)
3. \(A_1^2 \omega_1=A_2^2 \omega_2=A_3^2 \omega_3\)
4. \(A_1^2 \omega_1^2=A_2^2 \omega_2^2=A^2\)
Subtopic:  Simple Harmonic Motion |
 90%
Level 1: 80%+
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The equation of motion of a particle is \({d^2y \over dt^2}+Ky=0 \) where \(K\) is a positive constant. The time period of the motion is given by: 
1. \(2 \pi \over K\) 2. \(2 \pi K\)
3. \(2 \pi \over \sqrt{K}\) 4. \(2 \pi \sqrt{K}\)
Subtopic:  Simple Harmonic Motion |
 78%
Level 2: 60%+
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The kinetic energy of a particle executing SHM is \(16~\text{J}\) when it is in its mean position. If the amplitude of oscillations is \(25~\text{cm}\) and the mass of the particle is \(5.12~\text{kg}\), the time period of its oscillation will be:
1. \(\frac{\pi}{5}~\text{s}\)
2. \(2\pi ~\text{s}\)
3. \(20\pi ~\text{s}\)
4. \(5\pi ~\text{s}\)
Subtopic:  Energy of SHM |
 72%
Level 2: 60%+
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The period of oscillation of a simple pendulum of length \(L\) suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination \(\theta\), is given by:
1. \(2\pi\sqrt{\frac{L}{g\cos\theta}}\)
2. \(2\pi\sqrt{\frac{L}{g\sin\theta}}\)
3. \(2\pi\sqrt{\frac{L}{g}}\)
4. \(2\pi\sqrt{\frac{L}{g\tan\theta}}\)

Subtopic:  Angular SHM |
 60%
Level 2: 60%+
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