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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)

User HpsMouse
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1 Answer

2 votes

Answer:


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

Step-by-step 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{(K)/(m) }

The correct option is D

User Stuart Nichols
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