Physics, asked by Neel28vyas, 10 months ago

A wire of the length 0.5 m is stretched by a weight of 2 kg. If the mass per unit length of the wire is 1.96 ×

10-3

kg/m, find the fundamental frequency of the wire and frequency of its first overtone.​

Answers

Answered by sonuvuce
6

The fundamental frequency of wire is 100 Hz

Frequency of first overtone is 200 Hz

Explanation:

Length of the wire l=0.5 m

Weight attached =2 kg

Therefore, tension in the wire

T=2g N

Mass per unit length m=1.96\times 10^{-3} kg/m

We know that fundamental frequency of vibration is given by

\nu=\frac{1}{2l}\sqrt{\frac{T}{m}}

\implies \nu=\frac{1}{2\times0.5}\sqrt{\frac{2\times 9.8}{1.96\times 10^{-3}}}

\implies \nu=\sqrt{\frac{19.6}{1.96\times 10^{-3}}}

\implies \nu=\sqrt{\frac{1.96\times 10}{1.96\times 10^{-3}}}

\implies \nu=\sqrt{10^4} s⁻¹

\implies \nu=100 Hz

Frequency of first overtone

=2\nu

=2\times 100

=200 Hz

Hope this answer is helpful.

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Answered by Anonymous
1

Given :

The length of wire , l = 0.5 m

Weight  = 2 kg

Mass per unit , μ = 1.96 x 10 ^ (-3)

To find :

Fundamental frequency and frequency of the first overtone .

Solution :

Tension , T = mass * g

                   = 2 * 9.8

                   = 19.6 N

fundamental frequency , v = ( 1 / 2*l ) * √( T / μ )

                                            = ( 1 / 1 ) * √( 19.6 / 1.96 x 10 ^ (-3) )

                                            = √10000

                                            = 100 Hz

frequency of first overtone = 2 * fundamental frequency

                                             = 2 * 100 Hz

                                             = 200 Hz

The fundamental frequency is 100Hz and the frequency of the first overtone is 200 Hz .

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