Question 120867This question is from textbook Mathematics for Technicians
: Hello and good morning.
I am completely bamboozled with the following problem, and have therefore no idea as to how to go about solving it. Your assistance would be greatly appreciated. Thank you.
The frequency of vibration of a stretched wire when plucked varies inversely as its length and directly as the square root of its tension. If a wire that is 1.40 m long vibrates with a frequency of 376 per second (376 Hz), when its tension is 125 N:
A/ What will be its frenquency if its length is increased to 2.20 m and its tension increased to 234 N?
Thank you so very much.
Cheers
D
This question is from textbook Mathematics for Technicians
Found 3 solutions by josmiceli, scott8148, stanbon:
Answer by josmiceli(19441)   (Show Source):
Answer by scott8148(6628)  (Show Source):
You can put this solution on YOUR website! the formula would be f=k(sqrt(T))/L __ where f=freq, T=tension, L=length
__ and k is a constant (fudge factor) that accounts for the material of the wire, temperature, etc.
use the known parameters to find k, and then use k to find the second frequency
k=fL/(sqrt(T)) __ k=376*1.4/sqrt(125)
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! The frequency of vibration of a stretched wire when plucked varies inversely as its length and directly as the square root of its tension.
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EQUATION:
freq = k [sqrt(tension)/length]
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If a wire that is 1.40 m long vibrates with a frequency of 376 per second (376 Hz), when its tension is 125 N:
376 = k[sqrt(125)/1.4]
k = 1.4*376/sqrt(125) = 47.0826
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A/ What will be its frenquency if its length is increased to 2.20 m and its tension increased to 234 N?
freq = (47.0826)[sqrt(234)/2.2]
freq = 327.3755 Hz
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Cheers,
Stan H.
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