Question

3. A particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the origin, at t=0 s and moves to the right. The amplitude of its motion is 2.00 cm, and the frequency is 1.50 Hz. (a) Determine the position, velocity, and acceleration equations for this particle. (b) Determine the maximum speed of this particle and the first time it reaches this speed after t=0 s.

Answer 1

(a) As the particle is moving in a simple harmonic motion, the amplitude of the oscillation is 2 cm, frequency is 1.5 Hz and it starts from the origin(x=0), the equation of motion of the particle will be -

$x=2\sin (2\pi \times 1.5t)$

$\boldsymbol{\Rightarrow x=2\sin (3\pi t)}$ [where x is measured in cm]

To derive the velocity at any time, we must take derivative of x, so we will get -

$v=2\times 3\pi \cos (3\pi t)$

[where v is measured in cm/s]

v=67 cos(370)

Similarly to derive the acceleration at any time we must take derivative of the velocity, so we will get -

$a=-6\pi \times 3\pi \sin (3\pi t)$

$\boldsymbol{\Rightarrow a=-18\pi^{2} \sin (3\pi t)}$ [where a is measured in cm/s²]

(b) Clearly it can be seen from the expression of the velocity the maximum velocity of the particle is (6π).

[as maximum value of cos θ is 1].

If we put t=(1/3) sec in $v=6\pi \cos (3\pi t)$ , we get the speed |v|=6π. So after 0.33 seconds the particle will attain maximum velocity.