Question

A simple harmonic oscillator in the physics lab is measured to have an amplitude of 180.00 cm and a maximum speed of 350.0 cm/s. What is its speed when the displacement is 3.40 cm?

Answer 1

Amplitude = A
Max Speed = Aω
Displacement = x

Use Equations
x=A sin(ωt)
v=Aω cos(ωt)

Answer 2

i tried and couldnt solve it

Answer 3

Where did you got struck?

Answer 4

If you are struggling with trigonometry, use -
\[\sin \theta=\sqrt{1-\cos^{2} \theta}\]
It is not true in general, but will work in this case.

Answer 5

we have maxspeed = A \(\omega\) ---> \(\omega = \dfrac{maxspeed}{A}\)= \(\dfrac{350}{180}=1.94\)
And we have formula to calculate velocity at any time:
\[v = \sqrt{\omega(A^2 -x^2)}\]
just replace number in.

Answer 6

I didnt get the right answer with that either