The motion of the particle is given by the eq | vedclass

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

Question

(C) Given the equation of motion: $x = A \sin \omega t + B \cos \omega t$. To express this in the standard form of simple harmonic motion $x = R \sin(

Vedclass · Accepted Answer

simple harmonic with amplitude $(A^2+B^2)^{1/2}$

Vedclass · Answer

(C) Given the equation of motion: $x = A \sin \omega t + B \cos \omega t$. To express this in the standard form of simple harmonic motion $x = R \sin(\omega t + \phi)$,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 $\frac{A}{\sqrt{A^2 + B^2}} = \cos \phi$ and $\frac{B}{\sqrt{A^2 + B^2}} = \sin \phi$. Then,$x = \sqrt{A^2 + B^2} (\sin \omega t \cos \phi + \cos \omega t \sin \phi)$. Using the trigonometric identity $\sin(a+b) = \sin a \cos b + \cos a \sin b$,we get: $x = \sqrt{A^2 + B^2} \sin(\omega t + \phi)$. This is the standard equation of simple harmonic motion with amplitude $R = \sqrt{A^2 + B^2} = (A^2 + B^2)^{1/2}$.

The motion of the particle is given by the equation $x = A \sin \omega t + B \cos \omega t$. The motion of the particle is:

Similar Questions

Which of the following functions of time represent $(a)$ simple harmonic motion and $(b)$ periodic but not simple harmonic? Give the period for each case.
$(1)$ $\sin \omega t - \cos \omega t$
$(2)$ $\sin^2 \omega t$

Two equations of two $S.H.M.$ are $y = a\sin(\omega t - \alpha)$ and $y = b\cos(\omega t - \alpha)$. The phase difference between the two is .... $^\circ$.

Which of the following expressions corresponds to simple harmonic motion along a straight line,where $x$ is the displacement and $a, b, c$ are positive constants?

The function of time representing a simple harmonic motion with a period of $\frac{\pi}{\omega}$ is :

$A$ function is represented by the equation $y = A \cos \omega t \cos 2\omega t + A \sin \omega t \sin 2\omega t$. Then the nature of the function is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module