Question

The displacement (in m) of a particle of mass 100 g from
its equilibrium position is given by the equation,
y=0.05 sin 3 (5t +0.4)
(a) the time period of motion is
Sec
(b) the time period of motion is
Sec
(C) the maximum acceleration of the particle is 11.257
m/s2
(d) the force acting on the particle is zero when the
displacement is 0.05 m.​

Answer 1

Answer:

Time period of the motion is given as

$T = \frac{2\pi}{15} s$

Maximum acceleration of the particle is

$a = 11.25 m/s^2$

The force on the particle when it is at x = 0.05 is given as

$F = 1.125 N$

Explanation:

As we know that the equation of the motion is given as

$y = 0.05 sin3(5t + 0.4)$

here we know that the coefficient of the time is angular frequency

so we have

$\omega = 15$

so time period of the motion is given as

$T = \frac{2\pi}{15} s$

Now maximum acceleration of the particle is

$a = \omega^2 A$

$a = 15^2(0.05)$

$a = 11.25 m/s^2$

Also the force on the particle when it is at x = 0.05 is given as

$F = ma$

$F = 0.100 (11.25)$

$F = 1.125 N$

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Topic : SHM

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