Sonometer Experiment Problem
Question: In an experiment with a sonometer, when a mass of \(625\,\text{g}\) is attached to the string, it vibrates with a fundamental frequency of \(50\,\text{Hz}\). When a mass \(m\) is attached, the same string vibrates with a fundamental frequency of \(60\,\text{Hz}\). Find the value of \(m\) in grams.
(Assume that the tension in the string is produced by the weight of the attached mass and that the fundamental frequency is given by \[ f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}, \] where \(T = mg\) and all other factors remain constant.)
- (1) \(900\)
- (2) \(600\)
- (3) \(800\)
- (4) \(1200\)
Select the Correct Option
Step‑by‑Step Explanation
Step 1: The fundamental frequency of a vibrating string is given by: \[ f \propto \sqrt{T}, \] where the tension \(T\) is provided by the weight of the attached mass (\(T = mg\)). Since the string, its length \(L\), and the linear density \(\mu\) remain constant, we have: \[ f^2 \propto T. \]
Step 2: Let the first case (mass \(625\,\text{g}\) giving \(50\,\text{Hz}\)) correspond to tension \(T_1\) and the second case (mass \(m\) giving \(60\,\text{Hz}\)) correspond to tension \(T_2\). Since: \[ \frac{f_2^2}{f_1^2} = \frac{T_2}{T_1} \quad \text{or} \quad \frac{60^2}{50^2} = \frac{m}{0.625}, \] where mass is expressed in kilograms.
Step 3: Calculate the ratio: \[ \frac{60^2}{50^2} = \frac{3600}{2500} = \frac{36}{25}. \] Therefore, \[ \frac{36}{25} = \frac{m}{0.625}. \]
Step 4: Solve for \(m\): \[ m = 0.625 \times \frac{36}{25} = 0.625 \times 1.44 = 0.9\,\text{kg}. \] Converting to grams: \[ 0.9\,\text{kg} = 900\,\text{g}. \]
Final Answer: \(m = 900\,\text{g}\), which corresponds to option (1).
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