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### Frequency of potential energy in SHM = ?

A body is executing simple harmonic motion with frequency 'n', the frequency of its potential energy is:

(1) n                   (2) 2n
(3) 3n                 (4) 4n

Solution

In SHM, the potential energy is maximum at extreme positions and minimum at mean position. There are two extreme positions in one cycle and two mean positions. So, the time period should be halved or frequency should be doubled.

Alternate Solution

Potential Energy $U = \frac{1}{2}m{\omega ^2}{A^2}{\sin ^2}(\omega t + \phi )$

Or $U \propto {\sin ^2}(\omega t + \phi )$

Sin function alternates above and below x-axes whereas square of Sin function cannot be negative. So, the region below x-axis of Sin function is folded upwards above x-axes when it is squared. Thus the frequency is doubled or the time period is halved.

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$