1. A 0.33-kg mass is hanging from a spring with spring constant 12 N/m. Then the

Question

1. A 0.33-kg mass is hanging from a spring with spring constant 12 N/m. Then the mass is displaced from the equilibrium by 2.7 cm and let go. What is the resulting frequency of the oscillation?

2. A 0.8-kg mass is hanging from a spring with spring constant 19 N/m. Then the mass is displaced from the equilibrium by 2.3 cm and let go. What is the resulting period of the oscillation?

3.An ideal massless spring is hanging vertically. When a 0.54-kg mass is attached to the spring, the spring stretches by 3.7 cm before reaching a new equilibrium. Then the mass is displaced by 1.9 cm from the new equlibrium. What is the frequency of the resulting oscillation?

4. An ideal massless spring is hanging vertically. When an unknown mass is attached to the spring, the spring stretches by 4.8 cm before reaching a new equilibrium. Then the mass is displaced by 1.9 cm from the new equlibrium. What is the frequency of the resulting oscillation?

Explanation / Answer

Frequency = (1/(2pi))*sqrt(k/m)

= (1/(2pi))*sqrt(12/0.33)

= 0.959 Hz

Period = 2pi*sqrt(m/k)

= 2pi*sqrt(0.8/19)

= 1.289 sec

Here

Mg = kx

So

k/M = 9.8/0.037

= 264.864

As Frequency = (1/(2pi))*sqrt(k/m)

= (1/(2pi))*sqrt(264.864)

= 2.59 Hz

Here

Mg = kx

So

k/M = 9.8/0.048

= 204.1667

As Frequency  = (1/(2pi))*sqrt(k/m)

= (1/(2pi))*sqrt(204.1667)

= 2.274 Hz

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