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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Sound Waves

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Sound Waves

When the room temperature experiences a small variation from T to T...

When the room temperature experiences a small variation from TT to T+ΔTT + \Delta T, the fundamental frequency of an organ pipe shifts from ν\nu to ν+Δν\nu + \Delta\nu. Which of the following expressions correctly relates the fractional change in frequency to the fractional change in temperature?

Options

  1. AΔνν=2⋅ΔTT\dfrac{\Delta\nu}{\nu} = 2 \cdot \dfrac{\Delta T}{T}
  2. BΔνν=14⋅ΔTT\dfrac{\Delta\nu}{\nu} = \dfrac{1}{4} \cdot \dfrac{\Delta T}{T}
  3. CΔνν=ΔTT\dfrac{\Delta\nu}{\nu} = \dfrac{\Delta T}{T}
  4. DΔνν=12⋅ΔTT\dfrac{\Delta\nu}{\nu} = \dfrac{1}{2} \cdot \dfrac{\Delta T}{T}

Correct answer

D. Δνν=12⋅ΔTT\dfrac{\Delta\nu}{\nu} = \dfrac{1}{2} \cdot \dfrac{\Delta T}{T}

Step-by-step solution

The fundamental frequency of an organ pipe is given by ν=vλ\nu = \dfrac{v}{\lambda}, where vv is the speed of sound and λ\lambda is the wavelength.

The speed of sound in a gas at absolute temperature TT is given by v=γRTMv = \sqrt{\dfrac{\gamma R T}{M}}.

Thus, the fundamental frequency is directly proportional to the square root of the absolute temperature:

ν∝T\nu \propto \sqrt{T}

ν=kT1/2\nu = k T^{1/2}

Taking the natural logarithm on both sides:

ln⁡ν=ln⁡k+12ln⁡T\ln \nu = \ln k + \dfrac{1}{2} \ln T

Differentiating both sides for small variations:

Δνν=12ΔTT\dfrac{\Delta\nu}{\nu} = \dfrac{1}{2} \dfrac{\Delta T}{T}
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