Rankers Physics

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

A stretched string resonates with tuning fork frequency $512 \text{ Hz}$ when length of the string is $0.5 \text{ m}$. The length of the string required to vibrate resonantly with a tuning fork of frequency $256 \text{ Hz}$ would be: (1993)
$0.25 \text{ m}$
$0.5 \text{ m}$
$1 \text{ m}$
$2 \text{ m}$

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:

By the law of length for strings, $f_1 L_1 = f_2 L_2$. Substituting the values, $512 \times 0.5 = 256 \times L_2$. Solving for $L_2$ yields $1 \text{ m}$.

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