Question

A machine part is vibrating along the x-axis in simple harmonic motion with a period of 0.27 s and a range (from the maximum in one direction to the maximum in the other) of 3.0 cm. At time t = 0 it is at its central position and moving in the +x direction. What is its position when t = 55 s?

Answer 1

Answer:

$x = -1.437 cm$

Explanation:

The general equation for position of Simple harmonic motion is given as:

$x = A sin(\omega t)$          ........(1)

where,

x = Position of the wave

A = Amplitude of the wave

ω = Angular velocity

t = time

In this case, the amplitude is just half the range,

thus,

$A =\frac{3cm}{2}=1.5cm$  (Given range = 3cm)

A = 1.5 cm  

Now, The angular velocity is given as:

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

Where, T = time period of the wave =0.27s (given)

$\omega=\frac{2\pi}{0.27s}$

or

$\omega=23.27s^{-1}$

so, at time t = 55 s, the equation (1) becomes as:

$x = 1.5 sin(23.27\times 55)$

on solving the above equation we get,

$x = -1.437 cm$

here the negative sign depicts the position in the opposite direction of +x