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Waves question

2023 · 8 Apr · Shift 2 · Q60
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Waves question

2023 · 8 Apr · Shift 2 · Q60

JEE MainPhysicsWavesNumerical+4 / −1
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 ‾\underline{\hspace{2cm}}​ cm.
Numerical answer
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Correct answer: 60

  1. For a stretched string vibrating in the fundamental mode, f=v2Lf = \frac{v}{2L}f=2Lv​ where:

    • fff = fundamental frequency
    • vvv = wave speed in the string
    • LLL = length of the string

  2. Assuming the same string and same tension, the wave speed vvv remains constant. Hence, f∝1Lf \propto \frac{1}{L}f∝L1​ or f1L1=f2L2f_1 L_1 = f_2 L_2f1​L1​=f2​L2​

  3. Given: L1=90 cm,f1=120 Hz,f2=180 HzL_1 = 90\text{ cm}, \quad f_1 = 120\text{ Hz}, \quad f_2 = 180\text{ Hz}L1​=90 cm,f1​=120 Hz,f2​=180 Hz

  4. Using f1L1=f2L2f_1L_1 = f_2L_2f1​L1​=f2​L2​ we get 120×90=180×L2120 \times 90 = 180 \times L_2120×90=180×L2​

  5. Solve for L2L_2L2​: L2=120×90180L_2 = \frac{120 \times 90}{180}L2​=180120×90​ L2=2×903=60 cmL_2 = \frac{2 \times 90}{3} = 60\text{ cm}L2​=32×90​=60 cm

  6. Therefore, the required length of the string is 60 cm\boxed{60\text{ cm}}60 cm​

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