Simple Harmonic Motion (tuning fork)

[bebop721]

Q. the end of on of the prongs on a tuning fork that executes simple harmonic motion of frequency 1266 hz has an amplitude of 0.4944mm. find the speed of the end of the prong when the end has a displacement of 0.2037mm.

okay i know that you use the v(t) equations and all but where does the displacent fit in and what about "t" in those equations i am lost please someone help

 

[Doc Al]

You need to use both the displacement and velocity equations for simple harmonic motion. Review them here: http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html#c2"

 

[m2003]

I can help clarify the situation for you. Simple harmonic motion refers to the back and forth movement of an object at a constant frequency and amplitude. In this case, the object is a tuning fork with a frequency of 1266 Hz, meaning it completes 1266 cycles in one second.

The amplitude of the motion is the maximum displacement of the object from its equilibrium position. In this case, the amplitude is given as 0.4944mm. This means that the end of the prong moves back and forth by 0.4944mm from its resting position.

To find the speed of the end of the prong when it has a displacement of 0.2037mm, we can use the equation v = ω√(A^2 - x^2), where v is the speed, ω is the angular frequency (2πf), A is the amplitude, and x is the displacement.

Plugging in the values, we get v = (2π * 1266 Hz)√(0.4944mm^2 - 0.2037mm^2) = 2.522 mm/s. This means that when the end of the prong has a displacement of 0.2037mm, it is moving at a speed of 2.522 mm/s.

The "t" in the equation represents time, but it is not needed to solve this problem as we are given the frequency and amplitude. I hope this helps clarify the use of displacement in the v(t) equations for simple harmonic motion.