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 1/2 and 3/2 second with a speed of 3root2
m/s maximum speed of the particle (in m/s) will be​

Answer 1

Answer:

Maximum speed of the particle is 6 m/s

Explanation:

Time period of SHM is given as

$T = \frac{1}{2} + \frac{3}{2}$

$T = 2 s$

so we have angular frequency given as

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

$\omega = \frac{2\pi}{2} = \pi rad/s$

now let say particle start from position P, so equation of motion is given as

$x = A sin(\pi t + \phi)$

so here after 0.25 s the particle will reach to its maximum position so after this velocity will be zero

$v = A\pi cos(\pi t + \phi) = 0$ at t = 0.25 s

so we have

$\pi(0.25) + \phi = \frac{\pi}{2}$

so we have

$\phi = \frac{\pi}{4}$

now we know that speed at position P is 3root 2

so we have

$v = v_{max} cos(\pi t + \frac{\pi}{4})$

$3\sqrt2 = v_{max} cos(\frac{\pi}{4})$

$v_{max} = 6 m/s$

#Learn

Topic : SHM

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