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When tuning a violin, a string is tightened. The musical note being played on the string becomes higher in pitch.
A. True
B. False
C. Ambiguous
D. Data insufficient

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Last updated date: 13th Jun 2024
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Answer
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Hint: Think of which quantities that define the string are affected when we tune the violin. Also find an expression for frequency in terms of those quantities and see how the frequency is related to them. Also, remember higher the frequency, higher is the pitch.
Formula used:
Expression for frequency,
$f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu }}$

Complete answer:
In the question we are given a violin. When we are tuning a violin, we are actually tightening the string. We are also given a statement that: ‘The musical note being played on the string becomes higher in pitch.’ We are asked to find whether this statement is true or false.
If you ever had a practical experience with a musical instrument that has strings, you might be able to answer this directly. But we as physicists got other ways to find the answer but you should have a basic understanding on what actually happens while tuning a violin.
When you are tuning a violin (or guitar, or cello, or any other instrument that has strings), you are actually tightening the string of the violin. Doing so increases the tension in the string and also decreases the length of the string.
We know that frequency of a wave is given by,
$f=\dfrac{v}{\lambda }$
Where f is the frequency, v is the speed of the wave, and λ is the wavelength.
But we know that speed of a transverse wave on a stretched string is given by,
$v=\sqrt{\dfrac{T}{\mu }}$
Where, T is the tension on the string and μ is the linear mass density of the string.
Also, let L be the length of the string, for the lowest possible frequency of the system, that is, nodes of the vibrations produced are at the two ends of the string,
$L=\dfrac{\lambda }{2}$
$\Rightarrow \lambda =2L$
Therefore,
$\Rightarrow f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu }}$
When we tune a violin, tension (∝ frequency) is increased and length $\left( \propto \dfrac{1}{f} \right)$ is decreased, which clearly indicates that the frequency will increase as we tune and hence becomes higher in pitch.

So, the correct answer is “Option A”.

Note:
Shorter the stringer is, higher will be its frequency. Higher the tension in the string is, higher will be the frequency. Also, lighter the string is, higher the frequency is, that is, thicker string has lower frequency.