Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# 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. - Physics

Sum

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) = 0.0125, (b) t = 0.025.

#### Solution

Given are the equations of motion of a particle:
x1 = 2.0sin100 $\pi$t

$x_2 = 2 . 0\sin\left( 120\pi t + \frac{\pi}{3} \right)$

The Resultant displacement $\left( x \right)$ will be

x = x1 + x2

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

(a) At t = 0.0125 s

$x = 2\left[ \sin\left( 100\pi \times 0 . 0125 \right) + \sin\left( 120\pi \times 0 . 0125 + \frac{\pi}{3} \right) \right]$

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

$= 2\left[ \left( - 0 . 707 \right) + \left( - 0 . 5 \right) \right]$

$= 2 \times \left( - 1 . 207 \right) = - 2 . 41 cm$

(b) At t = 0.025 s

$x = 2\left[ \sin\left( 100 \pi \times 0 . 025 \right) + \sin\left( 120\pi \times 0 . 025 + \frac{\pi}{3} \right) \right]$ $= 2\left[ \sin\left( \frac{10\pi}{4} \right) + \sin\left( 3\pi + \frac{\pi}{3} \right) \right]$

$= 2\left[ 1 + \left( - 0 . 866 \right) \right]$

$= 2 \times \left( 0 . 134 \right) = 0 . 27 \text { cm }$

Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 12 Simple Harmonics Motion
Q 57 | Page 256