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Waves question

2010 · Q184
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Waves question

2010 · Q184

NEETPhysicsWavesMCQ+4 / −1
A tuning fork of frequency 512 Hz makes 4 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per sec when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was
  1. A
    510 Hz
  2. B
    514 Hz
  3. C
    516 Hz
  4. D
    508 Hz
View written solutionFree

Correct answer: D

Let the frequencies of tuning fork and piano string be υ1{\upsilon _1}υ1​ and υ2{\upsilon _2}υ2​ respectively.

∴\therefore∴ υ2=υ1±4=512 Hz±4{\upsilon _2} = {\upsilon _1} \pm 4 = 512\,Hz \pm 4υ2​=υ1​±4=512Hz±4

= 516 Hz or 508 Hz

AIPMT 2010 Prelims Physics - Waves Question 33 English Explanation


Increase in the tension of a piano string increases its frequency.

If υ2{\upsilon _2}υ2​ = 516 Hz, further increase in υ2{\upsilon _2}υ2​, resulted in an increase in the beat frequency. But this is not given in the question.

If υ2{\upsilon _2}υ2​ = 508 Hz, further increase in υ2{\upsilon _2}υ2​ resulted in decrease in the beat frequency. This is given in the question. When the beat frequency decreases to 2 beats per second. Therefore, the frequency of the piano string before increasing the tension was 508 Hz.

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