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Here is the question:

 

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A microphone is attached to a spring that is suspended from the ceiling, as the drawing indicates. Directly below on the floor is a stationary 400-Hz source of sound. The microphone vibrates up and down in simple harmonic motion with a period of 2.2 s. The difference between the maximum and minimum sound frequencies detected by the microphone is 2.3 Hz. Ignoring any reflections of sound in the room and using 343 m/s for the speed of sound, determine the amplitude of the simple harmonic motion.

 

Please tell me how to do this problem! I'm at 98.9 percent of my Webassign and all the Physics forums are slow to respond.

 

 

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nvm

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6 minutes ago, Johnmakuta said:

Come on, a formula, a webpage, anything!

 

in all seriousness, if you do not understand how to complete a question you were asked as homework you should just go talk to your teacher about it. It's his job to teach you this. 

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5 minutes ago, MimigaKing said:

in all seriousness, if you do not understand how to complete a question you were asked as homework you should just go talk to your teacher about it. It's his job to teach you this. 

I already did that with many other questions today and I an out of time for this one. Stop debating! Please only comment if you know Physics.

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Can we see the drawing

 

Nm I understand the question now.

 

The question is related to the doppler effect.

 

The microphone moves up and down every 2.2 seconds.

 

This means at some point the microphone is moving TOWARDS the sound source as well as AWAY from the microphone at a speed of V.

 

Using the original 400 hz as your original frequency f0, calculate what V must be to have a dopper shift of 2.3 hz.  (I dont remember the doppler equation)

 

Once you know V, calculate the amplitude the microphone must have a period of 2.2 hz

 

Hint:

 

k A^2 = m v^2 

Frequency^2 = k / m

Find frequency (rad/s) from a period of 2.2 sec

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