Question 458: Waves

Question

The fundamental frequency of sonometer wire increases by 9 Hz , if its tension is increased by $69 \%$, keeping the length constant. The frequency of the wire is

Accepted Answer

The fundamental frequency of a sonometer wire increases by 9 Hz when its tension is increased by 69%, with the length remaining unchanged. We need to determine the original frequency of the wire.
The frequency of vibration of a stretched string is determined by the formula:
$$ 
u = \frac{1}{2} \sqrt{\frac{T}{m}} $$
Where:
$ 
u $ is the frequency,
$ T $ is the tension in the string,
$ m $ is the mass per unit length,
$ I $ is the length of the string (though not included in this formula directly, itÃ¢ÂÂs relevant for the context of length being constant).
When the tension is increased by 69%, the new frequency becomes $ 
u + 9 $.
Let's calculate the new frequency using the increased tension:
$$ 
u^{\prime} = 
u + 9 = \frac{1}{2} \sqrt{\frac{T + (69/100)T}{m}} $$
On simplifying, we find the ratio of the old and new frequencies:
$$ \frac{
u}{
u^{\prime}} = \sqrt{\frac{T}{\frac{169}{100}T}} \Rightarrow \frac{
u}{
u+9} = \sqrt{\frac{100}{169}} = \frac{10}{13} $$
Solving the equation for the original frequency $ 
u $:
$$ 13 
u = 10 
u + 90 $$
This simplifies to:
$$ 
u = 30 	ext{ Hz} $$
Therefore, the frequency of the wire is 30 Hz.

Answer

42 Hz

Answer

24 Hz

Answer

30 Hz

Answer

36 Hz

Waves

The fundamental frequency of sonometer wire increases by 9 Hz , if its tension is increased by 69 \%, keeping the length constant. The frequency of the wire is

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