Question

A spring hangs from the ceiling. A block of 0.450 kg is tied to the other end of the spring. When released from rest, the block lowers 0.150 m before momentarily reaching rest, after which it moves upwards. (1) What is the spring constant, K? (2) Calculate the angular frequency of the vibrations of the block.

Answer 1

well, I am going to explain each and every step

Let me guess, you applied m * g = k * x. No wait maybe not, you have an 'h'. Anyways...

Apply conservation of energy.
(This is the third question on conservation of energy I am answering today...lol...)
Initial potential energy of spring is zero.
The block drops 0.15m. So the elongation in the srping is also 0.15m, as the block is attached to the spring.
So potential energy = 1/2 k x^2
where x = 0.15
PE = 1/2 k * (0.15)^2
PE = 1/2 k * 0.0225
PE = k * 0.01125
Change in PE = k * 0.01125 - 0 = k * 0.01125

Change in potential energy of spring = change in potential energy of block.
Since the block drops 0.15m, so the change is m * g * h, where h = 0.15m
PE = 0.45 * 9.8 * 0.15
Change in PE of spring = Change in PE of block
k * 0.01125 = 0.45 * 9.8 * 0.15
Solving for k, we get k = 58.8 N/m.

Time period T = (1 / 2*pi) * sqrt (m / k)
T = (1 / 2*pi) * sqrt (0.45 / 58.8)
T comes out to be equal to 0.014 s.
Angular frequency = 2*pi / T
Angular frequency = 2*pi / 0.014
Angular frequency = 11.4 rad/s

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