Q. 14.10 (b) 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

(b) at the maximum stretched position,

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

Answers (1)
S Sayak

Amplitude is A = 0.02 m

Time period is \\\omega

\\\omega =\sqrt{\frac{k}{m}}\\ \omega =\sqrt{\frac{1200}{3}}\\ \omega =20\ rad/s

(b) At t = 0 the mass is at the maximum stretched position.

x(0) = A

\phi =\frac{\pi }{2}

\\x(t)=0.02sin\left ( 20t + \frac{\pi }{2}\right )\\ x(t)=0.02cos(20t)

Here x is in metres and t is in seconds.

Exams
Articles
Questions