Question

A sewing machine needle moves in simple harmonic motion with a frequency of 2.5 Hz and an amplitude of 1.27 cm. (a) How long does it take the tip of the needle to move from the highest point to the lowest point in its travel? (b) How long does it take the needle tip to travel a total distance of 11.43 cm?

Answer 1

Answer:

The tip of the needle to move from the highest point to the lowest point in 0.4 sec and the needle tip to travel a total distance in 0.9 sec.

Explanation:

Given that,

Frequency = 2.5 Hz

Amplitude = 1.27 cm

(a). We need to calculate the time

The frequency is the reciprocal of the time.

$f=\dfrac{1}{T}$

$T=\dfrac{1}{f}$

$T=\dfrac{1}{2.5}$

$T=0.4\ sec$

The time taken from highest point to lowest point

$T'=\dfrac{T}{2}$

$T'=\dfrac{0.4}{2}$

$T'=0.2\ sec$

(b).  We need to calculate the time

The time taken in one cycle = 0.4 sec

The distance covered  in one sec= 4 times x amplitude

$d=4\times1.27$

$d=5.08\ m$

We need to calculate the speed

Using formula of speed

$v=\dfrac{5.08}{0.4}$

$v=12.7$

We need to calculate the time

$t=\dfrac{11.43}{12.7}$

$t= 0.9 sec$

Hence,  The tip of the needle to move from the highest point to the lowest point in 0.4 sec and the needle tip to travel a total distance in 0.9 sec.