Question

A Spring having with a spring constant 1200 N m⁻¹ is mounted on a horizontal table as shown in the figure. 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 x-axis. Give x as a function of time t for the oscillating mass if the moment we start the stopwatch (t = 0), the mass isa. at the mean positionb. at the maximum stretched position andc. at the maximum compressive positionIn what way do these functions for SHM differ from each other, in frequency and amplitude or the initial phase?

Answer 1

we know, motion of spring- block system is an example of simple harmonic motion. and equation of simple motion can be represented by $x=asin(\omega t\pm\Phi)$
where $\omega$ is angular frequency
a is amplitude of simple harmonic motion.
and $\Phi$ is phase constant.

for spring - block system,
$\omega^2=\frac{K}{m}$

here, m = 3kg [ as you forgot to write ]
displacement , a = 2cm
and spring constant, K = 1200N/m
now, $\omega=\sqrt{\frac{1200}{3}}$
$\omega=20rad/s$

so, equation of oscillation as function of time at mean position, $x=2sin20t$cm

b) At the maximum stretched position, the body will be at the extreme right position. The initial phase is π/2.
x = $2sin(\omega t+\pi/2)$ = 2cos20t cm

c) At the maximum compressed position, the body will be at the extreme left position. The initial phase is 3π/2.

x = $2sin(\omega t+3\pi/2)$= -2cos ωt = -2 cos 20t