Question

A 3.90 kg block hangs from a spring with spring constant 1980 N/m . The block is pulled down 5.40 cm from the equilibrium position and given an initial velocity of 1.80 m/s back toward equilibrium.

What is the frequency of the motion?

What is the amplitude?

What is the total mechanical energy of the motion?

Answer #1

for frequency use k = F/R = m w^2 where w is the angular
frequency in radians per second.

and k is the spring constant.

To convert w to frequency remember frequency = 2 /2Pi

So 1980 = 3.9 w^2

w = 22.532 and f = 22.532/2Pi = 35.39 Hz approx.

When the block is pulled down 0.054 m the system gains a net
energy = 1/2 k x^2

= 1/2 * 1980 * 0.054^2= 2.887 J

And it is given 1/2 m v^2 = 1/2 * 3.9 * 1.8 ^2 in kinetic. = 6.318 J

Add them up to get the total energy in the oscillation. = 6.318 + 2.887 J = 9.205 J

For the amplitude, at the maximum extension then 1/2 k x^2 = 9.205
J (total energy)

so 1/2*1980 x^2= 9.205

x = 0.096426

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