Question

A particle moves along the x-axis according to: $x=A[1+sin\omega t]$. What distance does it travel between t=0 and $t=2.5\pi /\omega$?

• A. 4A
• B. 6A
• C. 5A
• D. None

Answer 1

The correct option is C 5A

In the given SHM, particle starts from the extreme position. and the angular frequency of the given SHM is $\omega$.
So the time period of the given SHM will be $\frac{2\pi }{\omega }s$
We need to calculate the distance in $t=\frac{2.5\pi }{\omega }s=\frac{2\pi }{\omega }s+\frac{0.5\pi }{\omega }s$
or $t=(T+\frac{T}{4})s$
in one time period $\left(T\right)$ it will come back to its initial position so the distance traveled in time $Ts$ is $4A$ and after that it will reach to mean position in time $\frac{T}{4}s$, so the distance traveled in this time $\frac{T}{4}s$ will be $A$. So the total distance traveled in time $t=(T+\frac{T}{4})s$ will be equal to $4A+A=5A$