JEE MAIN - Physics (2021 - 18th March Evening Shift - No. 13)
The function of time representing a simple harmonic motion with a period of $${\pi \over \omega }$$ is :
cos($$\omega$$t) + cos(2$$\omega$$t) + cos(3$$\omega$$t)
sin2($$\omega$$t)
sin($$\omega$$t) + cos($$\omega$$t)
3cos$$\left( {{\pi \over 4} - 2\omega t} \right)$$
Giải thích
General equation of SHM
x = A sin($$\omega$$'t $$\pm$$ $$\phi$$)
We know, $$\omega$$ = $${{2\pi } \over T}$$
Given, $$T = {\pi \over \omega }$$
$$ \therefore $$ $$\omega$$' = $${{2\pi } \over {{\pi \over \omega }}}$$ = 2$$\omega$$
$$ \therefore $$ 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$$\omega$$.
x = A sin($$\omega$$'t $$\pm$$ $$\phi$$)
We know, $$\omega$$ = $${{2\pi } \over T}$$
Given, $$T = {\pi \over \omega }$$
$$ \therefore $$ $$\omega$$' = $${{2\pi } \over {{\pi \over \omega }}}$$ = 2$$\omega$$
$$ \therefore $$ 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$$\omega$$.