# The Frequency of Vibration of a String Depends on the Length L Between the Nodes, the Tension F in the String and Its Mass per Unit Length M. Guess the Expression for - Physics

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Sum

The frequency of vibration of a string depends on the length L between the nodes, the tension F in the string and its mass per unit length m. Guess the expression for its frequency from dimensional analysis.

#### Solution

Frequency, f $\propto$ LaFbmc
f = kLaFbmc   ...(1)
Dimension of [f] = [T−1]

Dimension of the right side components:
[L] = [L]
[F] = [MLT−2]
[m] = [ML−1]

Writing equation (1) in dimensional form, we get:
[T−1] = [L]a [MLT−2]b [ML−1]c
[M0L0T−1] = [Mb + c La + b − c T−2b]

Equating the dimensions of both sides, we get:
b + c = 0               ....(i)
a + b −c = 0      ....(ii)
−2b = −1              ....(iii)
Solving equations (i), (ii) and (iii), we get:

$a = - 1, b = \frac{1}{2} \text{ and } c = \left( - \frac{1}{2} \right)$
∴ Frequency, f = kL−1 F1/2m−1/2

$= \frac{k}{L} F^{1/2} m^\left( - \frac{1}{2} \right) = \frac{k}{L} \sqrt{\left( \frac{F}{m} \right)}$
$f = \frac{k}{L} \sqrt{\left( \frac{F}{m} \right)}$

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Chapter 1: Introduction to Physics - Exercise [Page 10]

#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 1 Introduction to Physics
Exercise | Q 17 | Page 10
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