A particle is performing SHM according to the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given equation for $SHM$ is $x = A \sin(\omega t + \phi)$,where $A = 3\, cm$ and $\omega = \frac{2\pi}{18} = \frac{\pi}{9}\, rad/s$.The time p

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(A) The given equation for $SHM$ is $x = A \sin(\omega t + \phi)$,where $A = 3\, cm$ and $\omega = \frac{2\pi}{18} = \frac{\pi}{9}\, rad/s$.The time period $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi/9} = 18\, s$.The total time given is $t = 36\, s$,which is equal to $2T$ (two complete oscillations).In one complete oscillation,the particle travels a distance of $4A$.Therefore,in $2$ complete oscillations,the distance travelled is $2 	imes 4A = 8A$.Substituting the value of $A = 3\, cm$,the distance travelled is $8 	imes 3 = 24\, cm$.

$A$ particle is performing $SHM$ according to the equation $x = (3\, cm) \sin \left( \frac{2\pi t}{18} + \frac{\pi}{6} \right)$ where $t$ is in seconds. The distance travelled by the particle in $36\, s$ is ..... $cm$.

Similar Questions

Define the amplitude of $SHM$ and draw two different $SHM$s in one figure having two different amplitudes.

What is a linear harmonic oscillator and a non-linear harmonic oscillator?

What is the amplitude of a particle executing uniform circular motion?

The equation of motion of a particle is $\frac{d^2y}{dt^2} + Ky = 0$,where $K$ is a positive constant. The time period of the motion is given by

What is a reference particle and a reference circle?

Vedclass Test Series

Exam Paper Generator

Online Exam Module