Question

A mass m is attached to a spring with a spring constant K. If the mass is set into simple harmonic motion by a displacement d from its equilibrium position, what would be the speed, v, of the mass when it returns to the equilibrium position?
A) v = sqrt(md/k)
B) v = sqrt(kd/m)
C) v = sqrtkd/mg)
D) v = d•sqrt(k/m)

Answer 1

Answer:

$(D), V =d.\sqrt{\frac{K}{m} }$

Explanation:

If the mass of the spring is set into simple harmonic motion at equilibrium position, the acceleration becomes zero and the speed will be maximum.

V = Aω

Also, if a mass m is attached to a spring with a spring constant K and  the mass is set into simple harmonic motion by a displacement d from its equilibrium position, then speed becomes;

V = dω

$V =d.\sqrt{\frac{K}{m} }$

The correct option is D