Question

the time period of oscillation of a particle whose position varies with time t as x=(sinwt + sin2Wt+sin4wt ) is

Answer 1

Answer:

See below.

Explanation:

We are given the displacement as a function of time as,

$x\left( t\right) =\sin \omega t+\sin 2\omega t+\sin 4\omega t$

If we are given a sum of functions which have different periods, we divide the LCM of the periods with the HCF of the periods.

We take T = 1 s and t = 1 s. Therefore, omega ( angular frequency ) is 2 pi.

The period of,

$\begin{aligned}\sin \omega t\rightarrow 2\pi \\ \sin 2 \omega t\rightarrow \pi \\ \sin 4\omega t\rightarrow \dfrac {\pi }{2}\end{aligned}$

For the time period of x{ t ],

$\begin{aligned}T'=LCM\left( 2\pi ,\dfrac {\pi }{2},\pi \right) \\ \overline {HCF\left( 2\pi ,\dfrac {\pi }{2},\pi \right) }\end{aligned}$

Which gives,

$T'=\dfrac {2\pi }{1}=2\pi$

Which is evident from the graph too. See the attached graph. The blue line is sin( x ) and the red line is the displacement function. Both have a period of 2 pi.

Answer 2

so the answer is 2pi/omega.Hope the attached pic helps you