Question 75: Oscillations

Question

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

Accepted Answer

General equation of SHM
x = A sin($\omega$'t $\pm$ $\phi$)
We know, $\omega$ = ${{2\pi } \over T}$
Given, $T = {\pi  \over \omega }$
$ 	herefore $ $\omega$' = ${{2\pi } \over {{\pi  \over \omega }}}$ = 2$\omega$
$ 	herefore $ Equation becomes,
x = a sin(2$\omega$t $\pm$ $\phi$)
Here coefficient of t is 2$\omega$.
you can see only option (D) has coefficient 2$\omega$.

Answer

cos($\omega$t) + cos(2$\omega$t) + cos(3$\omega$t)

Answer

sin
2
($\omega$t)

Answer

sin($\omega$t) + cos($\omega$t)

Answer

3cos$\left( {{\pi  \over 4} - 2\omega t} \right)$

Oscillations

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

Choose the correct answer: