A massless spring with no mass attached to it hangs from the ceiling. Its length

Question

A massless spring with no mass attached to it hangs from the ceiling. Its length is 0.2 meter. A mass M is now hung on the lower end of the spring. Imagine supporting the mass with your hand so that the spring remains relaxed (that is, not expanded or compressed), then suddenly remove your supporting hand. The mass and spring oscillate up and down. The lowest position of the mass during the oscillations is 0.1 m below the place it was resting when you supported it. (a) What is the frequency of oscillation? (b) What is the velocity when the mass is 0.05 m below its original resting place? A second mass of 0.3 kg is added to the first mass, making a total of M+0.3 kg When this system oscillates, it has half the frequency of the system with mass M alone. (c) What is the value of M? (d) Where is the new equilibrium position?

Explanation / Answer

We know for mass spring system

w=sqrt(K/M)

Frequency = w/2*pi

Since frequency is halved, so

Sqrt(M+0.3)=2*sqrt(M)

M=0.1 kg

(d)new equilibrium=0.6 m

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