Question

A 125 g pendulum bob hung on a string of length 35 cm has the same period as when the bob is hung from a spring and caused to oscillate. What is the spring’s elastic constant?a) 3.5 N/mb) 5.2 N/mc) 1.9 N/md) 27 N/m

Answer 1

k = 3.5 N/m

Step-by-step explanation:

It is given that the time period the bob in pendulum is the same as its time period in spring mass system:

`Time\ Period\ of\ Pendulum = Time\ Period\ of\ Spring-Mass\ System\\2\pi \sqrt{(l)/(g)} = 2\pi \sqrt{(m)/(k)`

`(l)/(g) = (m)/(k)\\\\ k = g(m)/(l)`

where,

k = spring constant = ?

g = acceleration due to gravity = 9.81 m/s²

m = mass of bob = 125 g = 0.125 kg

l = length of pendulum = 35 cm = 0.35 m

Therefore,

`k = (9.81\ m/s^2)((0.125\ kg)/(0.35\ m))\\\\`

k = 3.5 N/m