A guitar string of length 90 cm vibrates with a fundamental frequency of 120 Hz. The length of the string producing a fundamental frequency of 180 Hz will be _______ cm.
Correct Answer: 60
Solution and Explanation
The fundamental frequency (\( f \)) of a vibrating string is given by: \[ f = \frac{v}{2L}, \] where: \item \( f \) is the frequency, \item \( v \) is the wave velocity, \item \( L \) is the length of the string.
Step 1: Relation between frequencies and lengths. For the initial string length (\( L \)) and frequency (\( f_1 = 120 \, \text{Hz} \)): \[ f_1 = \frac{v}{2L}. \] For the new string length (\( L' \)) and frequency (\( f_2 = 180 \, \text{Hz} \)): \[ f_2 = \frac{v}{2L'}. \] Taking the ratio of the two frequencies: \[ \frac{f_2}{f_1} = \frac{L}{L'}. \] Substitute \( f_1 = 120 \, \text{Hz} \) and \( f_2 = 180 \, \text{Hz} \): \[ \frac{180}{120} = \frac{L}{L'}. \] Simplify: \[ \frac{3}{2} = \frac{L}{L'}. \] Rearrange to solve for \( L' \): \[ L' = \frac{2}{3} L. \]
Step 2: Substitute the initial length. Given \( L = 90 \, \text{cm} \): \[ L' = \frac{2}{3} \cdot 90 = 60 \, \text{cm}. \]
Final Answer: The new string length is: \[ \boxed{60 \, \text{cm}}. \]
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