Question

A particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the origin, at t = 0 and moves to the right. The amplitude of its motion is 2.00 cm, and the frequency is 1.50 Hz(a) show that the position of the particle is given by x = (2.00 cm) sin(3.00πt) (b) the maximum speed and the earliest time (t > 0) at which the particle has this speed, (c) the maximum
acceleration and the earliest time (t > 0) at which the particle has this acceleration, and (d) the total distance traveled between t = 0 and t = 1.00 s.

Answer 1

Answer:

(a). $x=2.00\sin(3.00\pi t)$, hence proved

(b). The maximum speed is 18.8 cm/s.

(c). The maximum  acceleration is 177.65 cm/s².

(d). The total distance is 12 cm.

Explanation:

Given that,

Amplitude = 2.00 cm

Frequency = 1.50 Hz

Given equation of position of the particle is

$x=2.00\ sin(3.00\pit)$

(a). show that the position of the particle is given by

$x=2.00\sin(3.00\pi t)$

We know the general equation of S.H.M

$x=A\sin(\omega t$...(I)

At t =0, x = 0

On differentiating equation (I)

$v=\dfrac{dx}{dt}$

$v=A\omega\cos(\omega t)$

At t = 0, the particle moving to the right

$V=A\omega$ > 0

Given statement is true.

The equation of position is

$x=A\sin(\omega t)$

here, $\Omega= 2\pi f$

Put the value in the equation

$x=2.00\sin(2\times1.50\pi t)$

$x=2.00\sin(3.00\pi t)$

Hence proved.

(b). We need to calculate the maximum speed

$V=A\omega\cos(\omega t)$....(II)

At t = 0,

$V_{max}=A\omega$

Put the value into the formula

$V_{max}=2.00\times2\pi\times1.50$

$V_{max}=6\pi$

$V_{max}=18.8\ cm/s$

(c). We need to calculate the maximum  acceleration

Using equation (II)

$V=A\omega\cos(\omega t)$

On differentiating

$a=\dfrac{dV}{dt}$

$a=-A\omega^2\sin(\omega t)$

$a_{max}\ when\ \sin(\omega t)\ is\ -1$

$a_{max}=-A\omega^2\times-1$

$a=A\omega^2$

$a_{max}=2\times(3\pi)^2\approx 177.65 cm/s^2$

(d). We need to calculate the total distance traveled between t = 0 and t = 1.00 s

Using equation (II)

$V=A\omega\cos(\omega t)$

On integration

$\int{V}=\int_{t}^{t'}{A\omega\cos(\omega t)}$

Put the vale into the formula

$\int{V}=\int_{0}^{1}{A\omega\cos(\omega t)}$

$D=\int_{0}^{1}|6\pi\cos\left(3\pi t\right)|dt$

$D=12\ cm$

Hence, (b). The maximum speed is 18.8 cm/s.

(c). The maximum  acceleration is 177.65 cm/s².

(d). The total distance is 12 cm.