MHT CET · Physics · Waves and Sound
A string in a musical instrument is \(50 \mathrm{~cm}\) long and its fundamental frequency is \(800 \mathrm{~Hz}\). Keeping the tension applied to the string same, the change in the length to produce sound note of fundamental frequency \(1000 \mathrm{~Hz}\) will be.
- A \(10 \mathrm{~cm}\)
- B \(20 \mathrm{~cm}\)
- C \(60 \mathrm{~cm}\)
- D \(40 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(D) \(40 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
For stretched string \(\mathrm{v} \propto \frac{1}{l}\)
\(\therefore \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{l_2}{l_1} \Rightarrow l_2=\frac{800}{1000} \times 50=40 \mathrm{~cm}\)
\(\therefore \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{l_2}{l_1} \Rightarrow l_2=\frac{800}{1000} \times 50=40 \mathrm{~cm}\)
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