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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Introduction to Physics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Introduction to Physics

The vibration frequency of a string depends on the length L between...

The vibration frequency of a string depends on the length LL between the nodes, the tension FF in the string, and its mass per unit length mm. Determine the expression for its frequency using dimensional analysis, assuming kk is a dimensionless constant.

Options

  1. AkLmF\dfrac{k}{L} \sqrt{\dfrac{m}{F}}
  2. BkLFmk L \sqrt{\dfrac{F}{m}}
  3. CkFmLk \sqrt{\dfrac{F}{mL}}
  4. DkLFm\dfrac{k}{L} \sqrt{\dfrac{F}{m}}

Correct answer

D. kLFm\dfrac{k}{L} \sqrt{\dfrac{F}{m}}

Step-by-step solution

Let the frequency ff be given by f=kLaFbmcf = k L^a F^b m^c.

Writing the dimensions of each physical quantity:
[f]=[T−1][f] = [T^{-1}]
[L]=[L][L] = [L]
[F]=[MLT−2][F] = [M L T^{-2}]
[m]=[ML−1][m] = [M L^{-1}]

Substituting the dimensions into the assumed relation:
[T−1]=[L]a[MLT−2]b[ML−1]c[T^{-1}] = [L]^a [M L T^{-2}]^b [M L^{-1}]^c
[T−1]=[M]b+c[L]a+b−c[T]−2b[T^{-1}] = [M]^{b+c} [L]^{a+b-c} [T]^{-2b}

Equating the powers of MM, LL, and TT on both sides:
b+c=0b + c = 0
a+b−c=0a + b - c = 0
−2b=−1-2b = -1

Solving these equations:
b=12b = \dfrac{1}{2}
c=−b=−12c = -b = -\dfrac{1}{2}
a=c−b=−12−12=−1a = c - b = -\dfrac{1}{2} - \dfrac{1}{2} = -1

Substituting the values of aa, bb, and cc back into the expression:
f=kL−1F1/2m−1/2=kLFmf = k L^{-1} F^{1/2} m^{-1/2} = \dfrac{k}{L} \sqrt{\dfrac{F}{m}}
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