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The prong of a tuning fork moves back and forth when it is set into vibration. The distance the prong moves between its extreme...Asked by Riya
The prong of a tuning fork moves back and forth when it is set into vibration. The distance the prong moves between its extreme positions is 2.28 mm. If the frequency of the tuning fork is 440.6 Hz, what are the maximum velocity and the maximum acceleration of the prong? Assume SHM.
I know that similar questions have been posted before, but some weren't answered and some where confusing, so if someone could help please!
I know that similar questions have been posted before, but some weren't answered and some where confusing, so if someone could help please!
Answers
Answered by
Riya
So what is Vm m/s
and Am m/s^2
Thank you!
and Am m/s^2
Thank you!
Answered by
Damon
x = 1.14 * 10^-3 sin 2 pi f t
in meters
f = 440.6
so w = 2 pi f = 2768 radians/s
x = 1.14 * 10^-3 sin 2768 t
v = 1.14*10^-3*2768 cos 2768 t
a = -1.14*10^-3*2768^2 sin 2768 t
v max = 1.14*10^-3*2768 = 3.16 m/s
a max = 2768 v max = 8736 m/s^2
in meters
f = 440.6
so w = 2 pi f = 2768 radians/s
x = 1.14 * 10^-3 sin 2768 t
v = 1.14*10^-3*2768 cos 2768 t
a = -1.14*10^-3*2768^2 sin 2768 t
v max = 1.14*10^-3*2768 = 3.16 m/s
a max = 2768 v max = 8736 m/s^2
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