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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान इयत्ता ११

A Wire Stretched Between Two Rigid Supports Vibrates in Its Fundamental Mode with a Frequency of 45 Hz. the Mass of the Wire is 3.5 × 10–2 Kg and Its Linear Mass Density is 4.0 × 10–2 Kg M–1. What is (A) the Speed of a Transverse Wave on the String, and (B) the Tension in the String?

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प्रश्न

A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10–2 kg and its linear mass density is 4.0 × 10–2 kg m–1. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?

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उत्तर १

a) Mass of the wire, m = 3.5 × 10–2 kg

Linear mass density, `mu = m/l = 4.0 xx 10^(-2) kg m^(-1)`

Frequency of vibration, ν = 45 Hz

:. Length of the wire, `l = m/mu = (3.5xx10^(-2))/(4.0xx10^(-2)) = 0.875 m`

The wavelength of the stationary wave (λ) is related to the length of the wire by the relation:

`lambda = (2l)/n`

Where n = number of nodes in the wire

For fundamental node, n = 1:

λ = 2l

λ = 2 × 0.875 = 1.75 m

The speed of the transverse wave in the string is given as:

= νλ= 45 × 1.75 = 78.75 m/s

b) The tension produced in the string is given by the relation:

v2µ

= (78.75)2 × 4.0 × 10–2 = 248.06 N

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उत्तर २

Here n= 45 Hz, M = `3.5 xx 10^(-2)` kg

Mass per unit length  = `m = 4.0 xx 10^(-2) kg m^(-1)`

`:. l = M/m = (3.5xx10^(-2))/(4.0xx10^(-2)) =  7/8`

As `l/2  =  lambda =  7/8 :. lambda =  7/4 m = 1.75 m`

a) The speed of the transverse wave,` v = vlambda = 45 xx 1.75 = 78.75 "m/s"`

b) As ` v = sqrt(T/m)`

`:. T = v^2xx m = (78.75)^2xx 4.0 xx 10^(-2) = 248.06 N`

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पाठ 14: Waves - Exercises [पृष्ठ ३८८]

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एनसीईआरटी Physics Part 1 and 2 [English] Class 11
पाठ 14 Waves
Exercises | Q 14 | पृष्ठ ३८८

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