waves and matter

Question

The point of a needle of a sewing machine acts as a simple harmonic oscillator. It moves in S.H.M along the y-axis with a frequency of 2.5Hz. At t=0, its position is 1.8cm and its velocity is 11cm/s. Find the magnitude of acceleration of the needle at t=0, to 3 s.f. in m/s²

Answer 1

See comments below for a correction

You could write the position function as y = Asin(ωt + φ) where A is the amplitude, ω is the frequency, and φ = the phase shift. That means, plugging in y = 1.8 and ω = 2.5 and t = 0 you get:

1.8 = Asin(2.5•0 + φ) or 1.8 = Asin(φ).

To find the velocity, we can take the derivative of the position equation y = Asin(2.5t + φ) to get:

v = y' = 2.5•Acos(2.5t + φ) and plugging in v = 11 at t = 0 we get:

11 = 2.5Acos(φ)

11/2.5 = Acos(φ)

4,4/cos(φ) = A

We can now plug "4.4/cos(φ)" into the first equation in place of "A" to get:

1.8 = Asin(φ)

1.8 = (4.4/cos(φ))sin(φ) and using the identity tan(φ) = sin(φ)/cos(φ) )we get:

1.8 = 4.4tan(φ) or tan(φ) = 1.8/4.4 or tan(φ) = 0.409091

Then φ = tan^-1(0.409091) meaning φ = 0.388319 radians

Using the first equation 1.8 = Asin(φ), we can plug in φ = 0.388319 and solve for A:

1.8 = Asin(0.388319) or A = 1.8/sin(0.388319) = 4.75395 so we can plug the values of φ and "A" into our equations of motion:

y = 4.75395sin(2.5t + 0.388319)

v = 11.88486cos(2.5t + 0.388319)

Taking the derivative of velocity:

a = v' = -2.5•11.88486sin(2.5t + 0.388319)

a = -29.71216sin(2.5t + 0.388319)

To find the acceleration at 0, plug in t = 0