Question

A particle is executing shm with time period t starting from mean position the time taken by it to complete 5/8 th oscillations is.......


(ͷϴ sϼαϻϻῖͷg ϼlϟ)​

Answer 1

Answer:

Time taken by it to complete 5/8 th oscillations is 7T/12

Explanation:

The angular frequency of oscillation is

$\omega =\frac{2\pi }{T}$

here the time required for one oscillation is called the time period T.

If the particle starts the oscillations from the origin and moves along the x-axis then the displacement-time relation is

$x=a sin\omega t$

Here  a is the amplitude

Now the time taken to complete 4/8 or half an oscillation is $T/2$

Now we need to do is determine the additional time required for completing 1/8 of an oscillation starting from the origin. which is given by

we know that 1/4 of an oscillation correspond to motion from the mean position to the extreme position is $x=a$. So 1/8 of oscillation will correspond to motion from the mean to $x=a/2$

$x=a sin\omega t\\\\\frac{a}{2} =a sin\omega t\\sin\omega t =\frac{1}{2}\\\\sin\omega t =sin\frac{\pi }{6}\\\\\omega t=\frac{\pi }{6}\\\frac{2\pi }{T} t=\frac{\pi }{6}\\t=\frac{T}{12}$

now time taken for 5/8 of an oscillation is

$=\frac{T}{2} +\frac{T}{12}\\=\frac{7T}{12}$