Question

A 51.0 g stone is attached to the bottom of a vertical spring and set vibrating. If the maximum speed of the stone is 16.0 cm/s and the period is 0.250 s, find the (a) spring constant of the spring, (b) amplitude of the motion, and (c) frequency of oscillation.

Answer 1

Answer:

A) 32.22 N/m b) 0.0156 m c) 4 Hz

Explanation:

Using Hooke's law;

T = 2π √m/k where m is mass of the body in kg and k is the force constant of the spring N/m and T is the period of vibration in s.

M = 51 g = 51 / 1000 in kg = 0.051kg

Make k subject of the formula

T/2π = √m / k

Square both sides

T^2 / 4π^2 = m/k

Cross multiply

K = 4 π^2 * m/T^2

K = 4 * 3.142 * 3.142 * 0.051/ 0.25^2= 32.22N/m

B) using Hooke's law;

F = k e where e is the maximum displacement of the spring from equilibrium point called amplitude

F= weight of the body = mass * acceleration due to gravity = 0.051*9.81

0.5 = 32.22 * e

e = 0.5/32.22 = 0.0156 m

C) frequency is the number of cycle completed in a second = 1 / period

F = 1 / 0.25 = 4Hz