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Karnataka Board PUCPUC Science Class 11

A 2⋅00 M-long Rope, Having a Mass of 80 G, is Fixed at One End and is Tied to a Light String at the Other End. the Tension in the String is 256 N. (A) Find the Frequencies of

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Question

A 2⋅00 m-long rope, having a mass of 80 g, is fixed at one end and is tied to a light string at the other end. The tension in the string is 256 N. (a) Find the frequencies of the fundamental and the first two overtones. (b) Find the wavelength in the fundamental and the first two overtones.

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Solution

Given:
Length of the long rope (L) = 2.00 m
Mass of the rope = 80 g = 0.08 kg
Tension (T) = 256 N
Linear mass density, 
\[= \frac{0 . 08}{2} = 0 . 04  \text{ kg/m }\]

\[Tension,   T = 256  N\] 

\[Wave  velocity,   v = \sqrt{\frac{T}{m}}\] 

\[ \Rightarrow v = \sqrt{\left( \frac{256}{0 . 04} \right)} = \frac{160}{2}\] 

\[ \Rightarrow v = 80  \text{ m/s }\]
For fundamental frequency:

\[L = \frac{\lambda}{4}\] 

\[ \Rightarrow \lambda = 4L = 4 \times 2 = 8  m\] 

\[ \Rightarrow f = \frac{v}{\lambda} = \frac{80}{8} = 10  \text{ Hz }\]
(a) The frequency overtones are given below:
\[\text{ 1st overtone } = 3f = 30 \text{ Hz }\]
\[\text{ 2nd overtone } = 5f = 50 \text{ Hz }\]
(b) \[\lambda = 4l = 4 \times 2 = 8  m\] 
\[\therefore \lambda_1 = \frac{v}{f_1} = \frac{80}{30} = 2 . 67 m\]
\[ \lambda_2 = \frac{v}{f_2} = \frac{80}{50} = 1 . 6 m\]
Hence, the wavelengths are 8 m, 2.67 m and 1.6 m, respectively.

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Chapter 15: Wave Motion and Waves on a String - Exercise [Page 327]

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HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 15 Wave Motion and Waves on a String
Exercise | Q 56 | Page 327

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