Question

A mass of 0.56 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by x(t) = (0.42 m)cos[(6 rad/s)t]. Determine the following. (a) amplitude of oscillation for the oscillating mass m (b) force constant for the spring N/m (c) position of the mass after it has been oscillating for one half a period m (d) position of the mass one-third of a period after it has been released m (e) time it takes the mass to get to the position x = −0.10 m after it has been released

Answer 1

Answer:

(a) 0.42 m

(b) 20.16 N/m

(c) - 0.42 m

(d) - 0.21 m

(e) 17.3 s

Solution:

As per the question:

Mass, m = 0.56 kg

x(t) = (0.42 m)cos[cos(6 rad/s)t]

Now,

The general eqn is:

$x(t) = Acos\omega t$

where

A = Amplitude

$\omega$ = angular frequency

t = time

Now, on comparing the given eqn with the general eqn:

(a) The amplitude of oscillation:

A = 0.42 m

(b) Spring constant k is given by:

$\omega = \sqrt{k}{m}$

$\omega^{2} = \frac{k}{m}$

Thus

$k = m\omega^{2} = 0.56\times 6^{2} = 20.16\ N/m$

(c) Position after one half period:

$x(t) = 0.42cos\pi = - 0.42\ m$

(d) After one third of the period:

$x(t) = 0.42cos(\frac{2\pi}{3}) = - 0.21\ m$

(e) Time taken to get at x = - 0.10 m:

$-0.10 = 0.42cos6t$

$6t = co^{- 1} \frac{- 0.10}{0.42}$

t = 17.3 s