Question

The time taken by a particle performing $SHM$ to pass from point $A$ to $B$ where its velocities are same is $2$ seconds.After another $2$ seconds it returns to $B$. The time period of oscillation is (in seconds)

• A. $2s$
• B. $8s$
• C. $6s$
• D. $4s$

Answer 1

The correct option is B $8s$

the Velocity $v$ in SHM is given by $v=\omega \sqrt{A^2-x^2}$. if at two points the velocity is equal that means, they are at equal distance from mean position. if one is at $x$ then another is at $-x$. So if it takes 2 seconds to reach from $A$ to $B$, it will take $1s$ for the particle to reach from $O$ to $B$. i.e.
$t_{OB}=1s$
Now it is given that it takes $2s$ to reach back at $B$.
$t_{BXB}=2s$
And it will take one more second to reach back at mean position.
$t_{BO}=1s$
So it will take $t_{OB}+t_{BXB}+t_{BO}=4s$ to complete the half cycle. so it will $8s$ to complete the full cycle.