Question

A particle is subjected to two simple harmonic motions given by x1 = 2.0 sin (100π t) and x2 = 2.0 sin (120 π t + π/3), where x is in centimeter and t in second. Find the displacement of the particle at (a) t = 0.0125, (b) t = 0.025.

Answer 1

Explanation:

• We can write the total displacement due to two simple harmonic motions as $x=x_{1}+x_{2}=2 \sin (100 \pi t)+2 \sin \left(120 \pi t+\frac{\pi}{3}\right)$  where,$x_1$  and $x_2$ are the given equations of motion of the particle.  Here x is measured in cm and t in second.  

• (a) We know that $x=2 \sin (1.25 \pi)+2 \sin \left(15 \pi+\frac{\pi}{3}\right)=2 \sin \left(\frac{\pi}{4}\right)-2 \sin \left(\frac{\pi}{6}\right)=>-\sqrt{2}-1 -2.414 \mathrm{cm}$ .  Therefore, we get the displacement of the particle at t=0.0125 s is =-2.414 cm.

• (b) Also we can write $x=2 \sin 25 \pi+2 \sin \left(3 \pi+\frac{\pi}{3}\right)=2 \sin \left(\frac{\pi}{3}\right)-2 \sin \left(\frac{\pi}{3}\right)=>\quad 2-\sqrt{3}= 0.27 \mathrm{cm}.$      Hence, we get the displacement of the particle at t=0.25 s is, 0.27 cm.