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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

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प्रश्न

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.

बेरीज

उत्तर

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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पाठ 15: Wave Motion and Waves on a String - Exercise [पृष्ठ ३२७]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
पाठ 15 Wave Motion and Waves on a String
Exercise | Q 56 | पृष्ठ ३२७

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