Two particles are executing Simple Harmonic Motion ($S$.$H$.$M$.). The equations of their motion are $y_1 = 10 \sin \left( \omega t + \frac{\pi}{4} \right)$ and $y_2 = 25 \sin \left( \omega t + \frac{\sqrt{3} \pi}{4} \right)$. What is the ratio of their amplitudes?

  • A
    $1:1$
  • B
    $2:5$
  • C
    $1:2$
  • D
    None of these

Explore More

Similar Questions

The radius of the circle,the period of revolution,the initial position,and the sense of revolution are indicated in the figure. The $y$-projection of the radius vector of the rotating particle $P$ is:

Match the following functions with their corresponding nature of motion, where $\omega$ is a constant:
List-$I$ List-$II$
$A$. $\sin^2 \omega t$ $I$. Periodic but not $SHM$ $(T = 2\pi/\omega)$
$B$. $\sin^3 \omega t$ $II$. Periodic but not $SHM$ $(T = \pi/\omega)$
$C$. $\sin \omega t + \cos \pi \omega t$ $III$. Non-periodic
$D$. $\cos \omega t + \cos 2\omega t$ $IV$. Periodic but not $SHM$ $(T = 2\pi/\omega)$

$A$ simple harmonic motion is represented by $y = 5(\sin 3\pi t + \sqrt{3} \cos 3\pi t) \ cm$. The amplitude and time period of the motion are:

The displacement of a particle is given by the relation $x = 4(\cos \pi t + \sin \pi t)$. The amplitude of the particle is

The motion of a particle is described by the equation $a = -bx$,where $a$ is the acceleration,$x$ is the displacement from the equilibrium position,and $b$ is a constant. The periodic time will be

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo