A 0.4-kg block is attached to a spring, pulled to the right and released from re

Question

A 0.4-kg block is attached to a spring, pulled to the right and released from rest, setting it into SHM on a
horizontal frictionless surface. The block's position is described by the equation x(t) = (17 cm)cos[(45.0 Hz)t],
where right is taken to be the positive direction.

1. Basics of the Simple Harmonic Oscillation.

(a) What are the amplitude and angular frequency of the block's oscillation?

A =  m
=  Hz

(b) What is the spring constant of the spring?

k =  N/m

(c) What is the mechanical energy of the block-spring system?

Esys =  J

Explanation / Answer

a) A = 17 cm ,  angular frequency = 45.0 Hz

b) W = sqrt (k/m)

=> k = mW^2 = 0.4* 45.0^2 = 810 N/m

c)mechanical energy of the block-spring system = 1/2KA^2 = 1/2*810*0.17^2 = 11.70 J

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