Question

A particle is subjected to two simple harmonic motions given by

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

where x is in centimeter and t in second. Find the displacement of the particle at

(a) t = 0.0125,

(b) t = 0.025

• A. 241 cm, -0.27 cm
• B. -241 cm, 0.27 cm
• C. 1.61 cm, -0.71 cm
• D. -1.61 cm, -0.71 cm

Answer 1

The correct option is B -241 cm, 0.27 cm

Both SHM's are along the x-axis and are

given $by{x}_1=2.0sin\left(100\pi t\right)cm$

$x_2=2.0sin(120\pi t+\frac{\pi }{3})$ cm

the resultant displacement will just be a sum of x1 and x2, as shown:

$x=x_1+x_2=2\left(sin\right(100\pi t)+sin(120\pi t+\frac{\pi }{3}\left)\right)$

putting values for t in the equation for x, we see:

x at t = 0.0125 s = -2.41 cm

x at t = 0.025 s = 0.27 cm