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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 - Physics

Sum

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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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 15 Wave Motion and Waves on a String
Q 56 | Page 327
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