Question

A mass is placed at the end of a spring. It has starting velocity of V & allowed to oscillate freely. If the mass has a starting velocity of 2V, what would the period be? Half as long, remains constant, 1/4 as long , 4 times as long, or 2 times as long?
​

Answer 1

Answer:

Equation for SHM can be written

V = w A cos w t        where w is the angular frequency and the velocity is a                                         maximum at t = 0

V1 = w1  A cos w1 t

V2 = w2 A cos w2 t

V2 / V1 = w2 / w1     since cos X t = 1 if t = zero

V2 / V1 = 2 pi f2 / (2 pi f1) = f2 / f1 = T1 / T2

If the velocity is twice as large the period will be 1/2 long