Question

A particle is executing SHM on a straight line. A and B are two points at which its velocity is zero. It passes through a certain point P(AP < PB) at successive intervals of 0.5 and 1.5 sec. with a speed of $3\sqrt{2}m/s$. Find the maximum speed (in m/s).

Answer 1

Time to go from P to A and back to P = 0.5s
Time to go from P to B and back to P = 1.5s
Here time period T = 1.5 to 0.5 = 2.0s
Now, angular velocity $\omega$ is given by,

$\omega =\frac{2\pi }{T}=\frac{2\pi }{2}=\pi \text{r}ad/s$
Since, the time to go from P to A and back to P = 0.5s. So, we can say,
Now let say particle start from position P, so equation of motion is given as:
$v=Asin(\pi t+\varphi )$
So here after 0.25 s the particle will reach to its maximum position so after this velocity will be zero.
$(\pi t+\varphi )=\pi /2$
So, $\varphi =\pi /4$

Now we know that speed at position P is $3\sqrt{2}$

So we have:

$v=v_{max}cos(\pi t+\varphi )$
$3\sqrt{2}=v_{max}cos\left(\varphi /4\right)$
$v_{max}=6m/s$