Question

I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block and it oscillates with a period of 0.13 s. What is the amplitude of oscillation

Answer 1

The amplitude of the oscillation is 2.82 cm

Step-by-step explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

`T = 2\pi \sqrt{(m)/(k) }`

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

`T = 2\pi \sqrt{(m)/(k) } \\\\k = (m*4\pi ^2)/(T^2) \\\\k = (4.1*4*(3.142^2))/((0.13^2)) \\\\k = 9580.088 \ N/m\\\\`

Now, determine the amplitude of oscillation, A;

`E = (1)/(2) kA^2`

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

`E = (1)/(2) kA^2\\\\2E = kA^2\\\\A^2 = (2E)/(k) \\\\A = \sqrt{(2E)/(k) } \\\\A = \sqrt{(2*3.8)/(9580.088) }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm`

Therefore, the amplitude of the oscillation is 2.82 cm