Question

A 140 g air-track glider is attached to a spring. The glider is pushed in 8.2 cm against the spring, then released. A student with a stopwatch finds that 14 oscillations take 14.0 s. Part A What is the spring constant?

Answer 1

Answer:

Explanation:

The period of oscillation is given as

T=2π√m/k

Making k subject of the formula

Square both sides of the equation

T²=4π²(m/k)

Cross multiply

T²k=4π²m

Then, divide through by T²

k=4π²m/T²

Where

k is spring constant

m is the mass of the bob

And T is the period of the oscillation

m=140g=0.14kg

14 oscillations takes 14 seconds

Then the period is

T=time/oscillation

T=14/14

T=1sec

Then,

k=4π²m/T²

k=4π²×0.14/1²

k=1.76N/m

Then, the spring constant is 1.76N/m