Question

A particle is subjected to two simple harmonic motions given by

$x_1=2.0sin\left(100\pi t\right)$ and $x_2=2.0sin(120\pi t+\pi /3)$,

where x is in centimeter and t in second. Find the displacement of the particle at (a) $t=0.0125,$ (b) $t=\mathrm{0.025.}$

Answer 1

$x_1=2sin100\pi t$

$x_2=2sin(120\pi t+\frac{\pi }{3})$

So, resultant displacement is given by

$x=x_1+x_2$

$=2\left[sin\right(100\pi t)+sin(120\pi t+\frac{\pi }{3})]$

(a) At $t=0.0125s$

$x=2\left[sin\right(100\pi \times 0.0125)+sin(120\pi \times 0.0125+\frac{\pi }{3})]$

$=2[sin\frac{5\pi }{4}+sin\left(\frac{3\pi }{2}\right)+\frac{\pi }{3}]$

$=2\left[\right(-0.707)+(-0.5\left)\right]$

$=2\times (-1.207)=-2.41cm$

(b) At $t=0.025s$

$x=2\left[sin\right(100\pi \times 0.025)+sin(120\pi \times 0.025+\frac{\pi }{3})]$

$=2[sin\left(\frac{10\pi }{4}\right)+sin(3\pi +\frac{\pi }{3})]$

$=2[1+(-0.866\left)\right]$

$=2\times \left(0.134\right)=0.27cm$