Simple Harmonic Motion (2 Viewers)

[locked.on]

Having some problems with the following:

The amplitude of a particle moving with SHM is 5 metres and the acceleration when 2 metres from the mean position is 4 m/s/s. Find the speed of the particle at the mean position and when it is 4 metres from the mean position.

 

[Drongoski]

acceleration (x double dot) = - n^2 x

ampl = a = 5

- 4 = - n^2 (2) ==> n = sqr(2)

v^2 = n^2(a^2 - x^2)

when x = 0: v^2 = 2(5^2 - 0^2) = 50
therefore speed = abs(sqr(50)) = 5sqr(2)

when x = 4

v^2 = 2(25 - 4^2) = 2x9

therefore speed = 3sqr(2)

hope answers right

 

[locked.on]

acceleration (x double dot) = - n^2 x

- 4 = - n^2 (2) ==> n = sqr(2)

How can acceleration here is negative?

Do we have to justify that it must be negative and never positive?
If so, how and why?

 

[Trebla]

Picture the wave diagram of x versus t. Assuming x = 0 is equilibrium, then at x = 2, it is above the equilibrium. This means it is on the section of the wave which is concave down (since it must reach a maximum before returning to x = 0), hence acceleration (which is the second derivative of x with respect to t) is negative.