Ncert Exercises & Solutions

14.10 In Exercise 14.9, let us take the position of mass when the spring is unstretched as x = 0, and the direction from left to right as the positive direction of the 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

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

The functions have the same frequency and amplitude, but different initial phases.
The amplitude of the mass, A = 2.0 cm

$Force constant of the spring, k = 1200 N m^{- 1}$
$Mass, m = 3 kg$
$Angular frequency of oscillation :$
$ω = \sqrt{\frac{k}{m}} = \sqrt{\frac{1200}{3}} = \sqrt{400} = 20 rad s^{- 1}$
$When the mass is at the mean position, initial phase is 0 .$
$Displacement, x = Asinωt = 2 \sin 20 t$
$At the maximum stretched position, the mass is toward the extreme right .$
$Hence, the initial phase is \frac{π}{2} .$
$Displacement, x = Asin (ωt + \frac{π}{2}) = 2 \sin (20 t + \frac{π}{2}) = 2 \cos 20 t$
$At the maximum compressed position, the mass is toward the extreme left .$
$Hence, the initial phase is \frac{3 π}{2} .$
$Displacement, x = Asin (ωt + \frac{3 π}{2}) = 2 \sin (20 t + \frac{3 π}{2}) = - 2 \cos 20 t$
$The functions have the same frequency (\frac{20}{2 π} Hz) and amplitude (2 cm), but different initial phases (0, \frac{π}{2}, \frac{3 π}{2}) .$

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