Rankers Physics

Standing Wave in String and Organ Pipe: Practice Problem & Solution

If the tension and diameter of a sonometer wire of fundamental frequency $n$ is doubled and density is halved then its fundamental frequency will become: (2001)
$\frac{n}{4}$
$\sqrt{2}n$
$n$
$\frac{n}{\sqrt{2}}$

Solution Explained:

To solve this problem, we apply the core principles of Standing Wave in String and Organ Pipe. Understanding the underlying formula is key to arriving at the correct answer below:

Fundamental frequency $n = \frac{1}{lD} \sqrt{\frac{T}{\pi \rho}}$. With new values $T' = 2T$, $D' = 2D$, and $\rho' = \rho/2$, the new frequency $n' = \frac{1}{2D} \sqrt{\frac{2T}{\pi (\rho/2)}} = \frac{1}{2D} \sqrt{\frac{4T}{\pi \rho}} = n$.

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