The displacement of an oscillator is given by | vedclass

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

Question

(C) Given the displacement equation: $x = a \sin \omega t + b \cos \omega t$. To simplify this,we multiply and divide by $\sqrt{a^2 + b^2}$: $x = \sqr

Vedclass · Accepted Answer

Motion is simple harmonic as well as periodic

Vedclass · Answer

(C) Given the displacement equation: $x = a \sin \omega t + b \cos \omega t$. To simplify this,we multiply and divide by $\sqrt{a^2 + b^2}$: $x = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \sin \omega t + \frac{b}{\sqrt{a^2 + b^2}} \cos \omega t \right)$. Let $\cos \phi = \frac{a}{\sqrt{a^2 + b^2}}$ and $\sin \phi = \frac{b}{\sqrt{a^2 + b^2}}$. Then the equation becomes: $x = \sqrt{a^2 + b^2} (\sin \omega t \cos \phi + \cos \omega t \sin \phi)$. Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get: $x = A' \sin(\omega t + \phi)$,where $A' = \sqrt{a^2 + b^2}$ is the resultant amplitude. Since the displacement is expressed in the form $x = A' \sin(\omega t + \phi)$,the motion is simple harmonic. Every simple harmonic motion is also periodic. Therefore,the motion is simple harmonic as well as periodic.

The displacement of an oscillator is given by $x = a \sin \omega t + b \cos \omega t$,where $a, b$ and $\omega$ are constants. Then:

Similar Questions

$A$ particle moves in $S.H.M.$ such that its acceleration is $a = -px$,where '$x$' is the displacement of the particle from the equilibrium position and '$p$' is a constant. The period of oscillation is

Consider a one-dimensional potential $V(x)$ as shown in the figure below. $A$ classical particle of mass $m$ moves under its influence and has total energy $E$ as shown below. The motion is

The displacement of a particle executing simple harmonic motion is given by
$y = A_{0} + A \sin \omega t + B \cos \omega t$
Then the amplitude of its oscillation is given by

What are the two basic characteristics of a simple harmonic motion?

The displacement of a particle in an oscillatory motion at a time $t$ is given by $x = 8 \sin \frac{\pi t}{4} \text{ cm}$. Calculate its displacement in the time interval $t = 0 \text{ s}$ to $t = 2 \text{ s}$. (in $\text{ cm}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module