Question

A machine part is undergoing SHM with a frequency of 5.00Hz and amplitude 1.80cm . How long does it take the part to gofrom x=0 to x= -1.80 cm?

Answer 1

To solve this exercise we will use the expression of the displacement in the aromatic movement in the two given points. Thus we will find the value of time as a function of frequency. In the end we will replace that value and find the time taken,

The expression for the displacement of the object in simple harmonic motion is

$x = Acos(2\pi ft)$

At the initial position the value of x is zero,

$0 = Acost (2\pi ft_1)$

$cost (2\pi ft_1) = 0$

$2\pi ft_1 = \frac{\pi}{2}$

$t _1 = \frac{1}{4f}$

Now the expression for the displacement of the object in simple harmonic motion is

$x = Acos (2\pi ft)$

At the final position the value of x is -1.8cm

Replacing we have that

$-1.8=1.8cos(2\pi ft_2)$

$-1 = cos(2\pi ft_2)$

$t_2 = \frac{1}{2f}$

The total change in time will be

$t = t_2-t_1$

$t = \frac{1}{2f}-\frac{1}{4f}$

$t = \frac{1}{4f}$

Replacing the value of the frequency we have that

$t = \frac{1}{4*5}$

$t = 0.05s$

Therefore the time taken for the displacement in the interval of x=0 to x=-1.8 is 0.05s