The $S.H.M.$ of a particle is given by the equation $x = 2 \sin \omega t + 4 \cos \omega t$. Its amplitude of oscillation is ........ units.

  • A
    $4$
  • B
    $2$
  • C
    $6$
  • D
    $2 \sqrt{5}$

Explore More

Similar Questions

One end of a rod of length $L$ is fixed to a point on the circumference of a wheel of radius $R$. The other end is sliding freely along a straight channel passing through the centre of the wheel as shown in the figure below. The wheel is rotating with a constant angular velocity $\omega$ about $O$. Taking $T = \frac{2 \pi}{\omega}$,the motion of the rod is

What is a reference particle and a reference circle?

$A$ stone is swinging in a horizontal circle $0.8 \, m$ in diameter at $30 \, rev/min$. $A$ distant horizontal light beam causes a shadow of the stone to be formed on a nearly vertical wall. The amplitude and period of the simple harmonic motion for the shadow of the stone are:

If the displacement $(x)$ and velocity $(v)$ of a particle executing simple harmonic motion are related through the expression $4v^2 = 25 - x^2$, then the time period is

Explain the concepts of a reference particle and a reference circle,and show that simple harmonic motion $(SHM)$ is the projection of uniform circular motion on a diameter of the reference circle.

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