Question

A 2.07-kg fish is attached to the lower end of an unstretched vertical spring and released. The fish drops 0.131 m before momentarily coming to rest. (a) What is the spring constant of the spring? (b) What is the period of the oscillations of the fish? ?

Answer 1

part a)

k = 310 N/m

part b)

T = 0.51 s

Step-by-step explanation:

Part A)

As per work energy theorem we have

Work done by gravity + work done by spring = change in kinetic energy

`mgx - (1)/(2)kx^2 = 0`

`(2.07)(9.8)(0.131) - (1)/(2)k(0.131)^2 = 0`

now we will have

`k = 310 N/m`

Part B)

Time period of oscillation is given as

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

`T = 2\pi\sqrt{(2.07)/(310)}`

`T = 0.51 s`