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 = (\pi)/(2)`

`t _1 = (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 = (1)/(2f)`

The total change in time will be

`t = t_2-t_1`

`t = (1)/(2f)-(1)/(4f)`

`t = (1)/(4f)`

Replacing the value of the frequency we have that

`t = (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