Question

a simple harmonic motion oscillator has spring constant k=5.0 N/m, amplitude A=10cm, and maximum speed 4.2m/s. what is the oscillator's speed when x=5.0cm?

Answer 1

i think i have to use \[v=\sqrt{k/m(A^2 -x^2)}\]

but i do not know how to solve for m to get V.

Answer 2

When X=A(amplitude)=10cm=0.1m then V=0.And when X=0 then V=maximum velocity=4.2m/sec
By the conservation of energy:
total mechanical energy(when X=A)=total mechanical energy(when V=maximum velocity)
i.e 1/2KA^2+0=0+1/2mV(maximum)^2...find m from this equation and put it in your equation V=sqrt(k/m(A^2-X^2)).

Answer 3

m=(k)*(A/Vmax)^2...
you can solve it also without formula V=sqrt(k/m(A^2-X^2)).
in simple, apply conservation of energy also here.
i.e total mechanical energy=1/2*k*A^2+0=0+1/2mVmax^2=1/2kx^2+1/2mv(required)^2...... apply this you can also get your answer in this way.