Question

Prove that the velocity of a particle in SHM at a distance √3/ 2 of it's amplitude is half of it's maximum velocity.​

Answer 1

Answer:

See below.

Explanation:

The general equation of the position of a body executing SHM is given by,

$x = A \ \sin( \omega t )$

Differentiating the above expression w.r.t to time gives,

$v = \frac{dx}{dt} = A \omega \cos( \omega t )$

The maximum velocity will be when,

$\cos(wt) = 1\\\implies v_{max} = A \omega$

Suppose the body is at a position that is root 3/2 of its amplitude. The time t will be,

$\frac{ \sqrt{3} }{2}A = A \ \sin(wt)\\\frac{ \sqrt{3} }{2} = \sin(wt)\\\implies wt = \frac{\pi}{3}\\\implies \frac{2 \pi}{T} t = \frac{\pi}{3}\\\implies t = \frac{T}{6}$

The velocity at this time will be,

$v_t = A \omega \cos( \omega \frac{T}{6} )\\v_t = A \omega \cos( \frac{2 \pi}{T} \frac{T}{6} )\\v_t = \frac{A \omega}{2}$

Hence the velocity at time t is half the maximum velocity.