Simple Harmonic Motion Maxiumum Velocity

[richardstan]

Hi, wondering if someone could help me with the following question.

A mass is hung from the lower end of a light vertical spring, whose upper end is fixed. The mass is pulled down and released and the time period of the oscillating system is measured. If the time period is 0.25s and the amplitude of oscillation is 30mm, calculate:

i.)the maximum speed
ii.)the maxiumum acceleration

I have an equation for the speed:

ds/dt = A 2(pi)f cos(2(Pi)ft)

i have a value for the frequency, is the maximum speed when the value of cos(2(Pi)ft) = 1?
Thanks
Richard.

 

[G01]

You got it.

 

[Moetasim]

Hello Richard,

Thank you for your question. In simple harmonic motion, the maximum velocity occurs when the displacement is at its maximum value, which is the amplitude A. This means that the maximum speed of the mass is equal to the amplitude multiplied by the frequency: vmax = A * f.

In this case, the amplitude is given as 30mm and the frequency can be calculated using the formula f = 1/T, where T is the time period of 0.25s. Therefore, the maximum speed would be vmax = (30mm) * (1/0.25s) = 120mm/s.

To calculate the maximum acceleration, we can use the formula a = -(2*pi*f)^2 * x, where x is the displacement. In this case, the maximum acceleration would be amax = -(2*pi*(1/0.25s))^2 * (30mm) = -7539.82 mm/s^2.

I hope this helps answer your question. Let me know if you have any further questions. Good luck with your calculations!

Best,