# The potential energy of a particle oscillating along the x-axis is given as U = 20+ (x–2)2 where U is in joules and x in meters. The total mechanical energy of the particle is 36 J. The maximum kinetic energy of the particle will be: 1. 24 J  2. 36 J 3. 16 J  4. 20 J

Subtopic:  Energy of SHM |
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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 |
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Kinetic energy of a particle executing simple harmonic motion in straight line is $$pv^2$$ and potential energy is $$qx^2,$$ where $$v$$ is speed at distance $$x$$ from the mean position. The time period of the SHM is given by the expression:

1. $2\mathrm{\pi }\sqrt{\frac{\mathrm{q}}{\mathrm{p}}}$

2. $2\mathrm{\pi }\sqrt{\frac{p}{q}}$

3. $2\mathrm{\pi }\sqrt{\frac{\mathrm{q}}{\mathrm{p}+\mathrm{q}}}$

4. $2\mathrm{\pi }\sqrt{\frac{p}{p+q}}$

Subtopic:  Energy of SHM |
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A particle is executing SHM according to y = a cos$\mathrm{\omega t}$. Then, which of the following graphs represent variations of potential energy?

1. I and III

2. II and IV

3. II and III

4. I and IV

Subtopic:  Energy of SHM |
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Two springs, of force constants k1 and k2 are connected to a mass m as shown in the figure. The frequency of oscillation of the mass is f. If both k1 and k2 are made four times their original values, the frequency of oscillation will become:

 1 2f 2 f/2 3 f/4 4 4f
Subtopic:  Combination of Springs |
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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 |
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The amplitude and the time period in an S.H.M. are 0.5 cm and 0.4 sec respectively. If the initial phase is $\mathrm{\pi }/2$ radian, then the equation of S.H.M. will be:

1. $\mathrm{y}=0.5\mathrm{sin}5\mathrm{\pi t}$

2. $\mathrm{y}=0.5\mathrm{sin}4\mathrm{\pi t}$

3. $\mathrm{y}=0.5\mathrm{sin}2.5\mathrm{\pi t}$

4. $\mathrm{y}=0.5\mathrm{cos}5\mathrm{\pi t}$

Subtopic:  Linear SHM |
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In simple harmonic motion, the ratio of acceleration of the particle to its displacement at any time is a measure of:

 1 Spring constant 2 Angular frequency 3 (Angular frequency)2 4 Restoring force
Subtopic:  Simple Harmonic Motion |
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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 |
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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. $$0$$ and $$2K_0$$
2. $$\frac{K_0}{2}$$ and $$K_0$$
3. $$K_0$$ and $$2K_0$$
4. $$K_0$$ and $$K_0$$

Subtopic:  Energy of SHM |
62%
From NCERT
AIPMT - 2007
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