Below is the correct question
sewing machine needle moves up and down in simple harmonic motion with an amplitude of 0.0127 m and a frequency of 2.55 Hz.
(a) What is the maximum speed of the needle
(b) What is the maximum acceleration of the needle
Answer:
(a) [tex]V_{speed}=0.2034m/s[/tex]
(b) [tex]a_{acceleration}=0.002583m/s^{2}[/tex]
Explanation:
Given data
Frequency f=2.55 Hz
Amplitude A=0.0127 m
To find
(a) Speed
(b) Acceleration
Solution
For (a) the maximum speed of the needle
We can find the speed as:
[tex]V_{speed}={A}{2(\pi)f }\\V_{speed}=0.0127m*2(\pi) (2.55Hz)}\\ V_{speed}=0.2034m/s[/tex]
For (b) the maximum acceleration of the needle
[tex]a_{acceleration}=A^{2}*2(\pi)f\\a_{acceleration}=(0.0127m)^{2}*(2\pi )(2.55Hz)\\ a_{acceleration}=0.002583m/s^{2}[/tex]
If a sewing machine needle moves up and down in simple harmonic motion with an amplitude of 0.0127 m and a frequency of 2.55 Hz then the needle move in one period = 0.0508 m
If the displacement of a particle undergoing simple harmonic motion of amplitude A at a time is [tex]→x=Asin[/tex]ω[tex]t^i[/tex]
= [tex](+A^i)+ (-2A^i)+(+A^i)[/tex]
= [tex]0^i.[/tex]
Given:
Amplitude = 0.0127 m
frequency = 2.55 Hz
Solution:
To find the displacement of the needle in one period we need to put value in the formula:
The total distance traveled = 4A
= 4*0.0127 m
= 0.0508 m
Thus, If a sewing machine needle moves up and down in simple harmonic motion with an amplitude of 0.0127 m and a frequency of 2.55 Hz then the needle move in one period = 0.0508 m
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