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Varvara68 [4.7K]
3 years ago
11

Two violinists are trying to tune their instruments in an orchestra. One is producing the desired frequency of 440.0 hz. The oth

er is producing a frequency of 448.4 hz. By what percentage should the out-of-tune musician change the tension in his string to bring his instrument into tune at 440.0 hz?
Physics
1 answer:
Katyanochek1 [597]3 years ago
4 0

Answer:

Percentage change in tension is 3.8%

Explanation:

We have given initially frequency f_1 = 440 Hz

Let tension in the string at this frequency is T_1

Now second frequency is f_2=448.4Hz

Frequency in string is given by

f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}

From the relation we can say that

\frac{f_1}{f_2}=\sqrt{\frac{T_1}{T_2}}

\frac{440}{448.4}=\sqrt{\frac{T_1}{T_2}}

{\frac{T_2}{T_1}}=1.038

Percentage change in tension is equal to

=\frac{T_2-T_1}{T_1}=\frac{T_2}{T_1}-1=(1.038-1)\times 100=3.8 %

So percentage change in tension is 3.8%

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Explanation:

Part (a)

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