Can anybody help with a question about Simple harmonic motion?

An object attached to the end of the spring is undergoing simple harmonic motion with an amplitude of 4.5*10^-2 m and the period of the motion is 0.90s. What is the maximum speed of the object?

Can anybody help with a question about Simple harmonic motion?the position as a function of time is



(4.5e-2 m)*cos(w*t + some phase)



where w is the angular frequency. w can be calculated from the period as



w=2*pi/T = 2*pi/(0.9s)



the velocity is the derivative of the position:



-(4.5e-2)*w*sin(w*t+ phase)



so the max speed is just the amplitude of the velocity



(4.5e-2 m)*w = (4.5e-2 m)*2*pi/(0.9s)



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if you're unfamiliar with calculus and don't know what "derivative" means, here's how you proceed instead:



the total energy is the potential+kinetic parts:



E = (k/2)*x^2 + (m/2)*v^2



where x is the position and v is the velocity. we know the position, we're trying to solve for the amplitude of the velocity.



v^2 = 2*E/m - (k/m)*x^2



notice that at a certain time all the energy is potential, so E is simply (k/2)*(4.5e-2 m)^2:



v^2 = (k/m)*((4.5e-2 m)^2)*[1 - ( cos(w*t + phase) )^2]

v^2 = (k/m)*((4.5e-2 m)^2)*[sin(w*t + phase)]^2



finally, notice that k/m = w^2.



|v| = w*(4.5e-2 m)*|sin(w*t + phase)|, and the amplitude of the velocity is just



w*(4.5e-2 m) = 2*pi*(4.5e-2 m)/(0.9s), as we found before.Can anybody help with a question about Simple harmonic motion?6 miles per second.Can anybody help with a question about Simple harmonic motion?ask simple simon.....he says.........................