Question

An oscillator consists of a block of mass 0.500 kg connected to a spring. When set into oscillation with amplitude 35.0 cm, the oscillator repeats its motion every 0.500 s. Find the (a) period, (b) frequency, (c) angular frequency, (d) spring constant, (e) maximum speed, and (f) magnitude of the maximum force on the block from the spring.

Answer 1

a. t=0.5s

b. f=2Hz

c. w =4π

d. k = 78.95 N/m

e. v = 4.39 m/s

f. Fs=4.835N

Step-by-step explanation:

a. The period of the motion is

t= 0.5 s

b. Frequency

f =1 / t

f = 1 / 0.5 = 2 Hz

c. Angular frequency

w = 2π*f

w = 2π*2Hz= 4π

d. The spring constant is

f = 1/2π * √k/m

k = (2π * f)^2*m =(2π*2Hz)^2 * 0.5 kg

k = 78.95 N/m

e. Maximum speed

Kinetic energy and spring energy to find the maximum speed

1/2 *K*x^2=1/2*m*v^2

1/2 * 78.95N/m*(0.35m)^2=1/2*0.5kg*v^2

Solve to v'

v = 4.39 m/s

f. The maximum force on the block from the spring

Fs= 1/2*K*x^2

Fs=1/2*78.95N/m*(0.35m)^2

Fs=4.835N