The graph of velocity versus time for a simple harmonic oscillator is given belo

Question

The graph of velocity versus time for a simple harmonic oscillator is given below. Determine: the amplitude frequency and angular frequency of the motion; the position at t = 0; the position, velocity and acceleration as functions of time; assuming the oscillator is a 2.00 kg block attached to a horizontal spring on a frictionless surface, find the spring constant k; the total energy stored in the system; the kinetic energy and potential energy as functions of time; the maximum force experienced by the block.

Explanation / Answer

we can see it repeats twice by 10 seconds so T= 5
so =2/5

the max velocity is 7

max velocity = A

a) so A = .07/ (2/5) = .35/2

b) if v=0, we are at max position so at t=0, x= .35/2

c) x = -.35/2 cos ( 2/5 t)

v= .07 sin(2/5 t)

a= .14/5 cos(2/5 t)

d) =sqrt(k/m)

k= m^2= 2* (2/5)^2 = 8 ^2/25

e) at max velocity x=0 so E= 1/2 m vmax^2 = 1/2 * 2 * .07^2 = .07^2

f) KE = 1/2 mv^2 = 1/2 * 2 * v^2 = .07^2 sin(2/5 t) ^2

PE = 1/2 k x^2 = 4 ^2/25 (.35/2 cos ( 2/5 t))^2

g) F = kx

so F max = k xmax = 8 ^2/25* .35/2

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