wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In Exercise 14.9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is

(a) at the mean position,

(b) at the maximum stretched position, and

(c) at the maximum compressed position.

In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Open in App
Solution

(a) x = 2sin 20t

(b) x = 2cos 20t

(c) x = –2cos 20t

The functions have the same frequency and amplitude, but different initial phases.

Distance travelled by the mass sideways, A = 2.0 cm

Force constant of the spring, k = 1200 N m–1

Mass, m = 3 kg

Angular frequency of oscillation:

= 20 rad s–1

(a) When the mass is at the mean position, initial phase is 0.

Displacement, x = Asin ωt

= 2sin 20t

(b) At the maximum stretched position, the mass is toward the extreme right. Hence, the initial phase is.

Displacement,

= 2cos 20t

(c) At the maximum compressed position, the mass is toward the extreme left. Hence, the initial phase is.

Displacement,

= –2cos 20t

The functions have the same frequency and amplitude (2 cm), but different initial phases.


flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Applying SHM
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon