The assistant adjusts the tension in the same piano string, and a beat frequency

Question

The assistant adjusts the tension in the same piano string, and a beat frequency of 4.00 Hz is heard when the note and the tuning fork are struck at the same time. (a) Find the two possible frequencies of the string. flower = Hz fhigher = Hz

(b) Assume the actual string frequency is the higher frequency. If the piano tuner runs away from the piano at 3.30 m/s while holding the vibrating tuning fork, what beat frequency does he hear? f = Hz (c) What beat frequency does the assistant on the bench hear? Use 343 m/s for the speed of sound. f = Hz

Explanation / Answer

a) n1= frequancy of piano string

n2= frequancy of tunning fork

In order to solve this problem we need to determine the frequancy of tunning fork. If tunning fork frequancy is given then we can solve it further. Let the frequancy of tunning fork is meaured to be x. Then beat frequancy n is given a s

n = n2-n1

8 = x - n1

n 1= x - 8

So, f lower would be = f lower = n1= x -8Hz

If   

n1- n2= 8

Then f higer = n1= n2+ 8 = x + 8

v) The change in beat frequancy would be

By Doppler Shift

n1/ =(( V + Vo ) / V)) n1

The value of n1 is

n1 = x + 8

n1/ = ((V + V0) / V)) (x+8)

n1/ =(( 343 + Vo) / 343)) (x + 8)

Now there are two unknowns , if we know the value of x , then and only we can proceed further with calculation.

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