Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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The length of the wire shown in figure (15-E8) between the pulley is 1⋅5 m and its mass is 12⋅0 g. Find the frequency of vibration with which the wire vibrates in two - Physics

Sum

The length of the wire shown in figure between the pulley is 1⋅5 m and its mass is 12⋅0 g. Find the frequency of vibration with which the wire vibrates in two loops leaving the middle point of the wire between the pulleys at rest.

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Solution

Given:
Length of the wire between two pulleys (L) = 1.5 m
Mass of the wire = 12 gm

\[\text{ Mass  per  unit   length,}   m = \frac{12}{1 . 5}  g/m\] 

\[ = 8 \times  {10}^{- 3}   kg/m\]

\[Tension  in  the   wire,   T = 9 \times g\] 

\[ = 90  \text{ N }\]

Fundamental frequency is given by:
\[f_0  = \frac{1}{2L}  \sqrt{\left( \frac{T}{m} \right)}\]
For second harmonic (when two loops are produced):

\[f_1  = 2 f_0  = \frac{1}{1 . 5}  \sqrt{\left( \frac{90}{8} \times {10}^{- 3} \right)}\] 

\[  = \frac{\left( 106 . 06 \right)}{1 . 5}\] 

\[ = 70 . 7  Hz \approx 70  Hz\]

Concept: Wave Motion
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 15 Wave Motion and Waves on a String
Q 39 | Page 326
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