Question 99: 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

sin2($$\omega$$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: