-1
$\begingroup$

I have the question "A sewing machine needle moves up and down through a total vertical distance of 2.0cm. The frequency of the oscillation is 2.4 Hz. Assuming the motion is SHM (simple harmonic motion) calculate:

(a) The amplitude of the motion,

(b) The maximum acceleration of the needle."

For (a) the amplitude I got 0.02 m.

For (b) the maximum acceleration of the needle I used the equation:

a = w$^2$A

Therefore, a = (2$\pi$ 2.4)$^2$ x 0.02

Therefore, a = 4.55 ms$^-2$.

Are my answers correct?

1 Answers 1

1

The sewing machine needle is moving through a total vertical distance of $2\,\mathrm{cm}$. For simple harmonic motion, the amplitude is half this. To see this, note the total vertical distance is the distance between peaks in $\sin(\cdot )$. The amplitude is simply the height of one peak.

  • 0
    I got amplitude = 0.01 m and so the maximum acceleration is now 2.27 ms$^-2$ is this now correct ? – 2017-01-06
  • 1
    @Dan Yes, that is correct. – 2017-01-06