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?

Answer 1

Let standard equation of SHM is
X(t) = Asin( wt + ∅)

(a) when x = 0 , t = 0
0 = Asin(w×0 + ∅)
∅ = 0
So, required function is x(t) = Asinwt
w = √{K/m}
= √{ 1200/3} = 20 rad/s
So, x(t) = Asin20t = 2sin20t

(b) when t = 0, x = +A
A = Asin(w× 0 + ∅)
1 = sin∅
∅ = π/2
So, equation is x(t) = Asin(wt + π/2)
x(t) = 2sin(20t + π/2)
X(t) = 2cos20t

(c) when t = 0 , x = -A
x(t) = Asin(wt + ∅)
-A = Asin(w × 0 + ∅)
∅ = -π/2
So, equation is x(t) = Asin(20t - π/2)
x(t) = -2cos20t