A wire of length 2.00 m is subjected to a tension of 160 N . Determine the linear mass density of the wire if its fundamental frequency of vibration is 100 Hz .

Correct answer: A. 1.0 × 10^ -3 kg/m The fundamental frequency of a stretched wire is given by: f = (1)/(2L) (T)/(μ) where L is the length, T is the tension, and μ is the linear mass density. Rearranging the formula to solve for μ : μ = (T)/((2Lf)^2) Substituting the given values L = 2.00 m , T = 160 N , and f = 100 Hz : μ = (160)/((2 × 2.00 × 100)^2) μ = (160)/((400)^2) μ = (160)/(160000) μ = 1.0 × 10^ -3 kg/m

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Wave Motion and Waves on a String

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Wave Motion and Waves on a String

A wire of length 2.00 m is subjected to a tension of 160 N ....

A wire of length

2.00 \text{ m}

is subjected to a tension of

160 \text{ N}

. Determine the linear mass density of the wire if its fundamental frequency of vibration is

100 \text{ Hz}

Options

A $1.0 \times 10^{-3} \text{ kg/m}$
B $8.0 \times 10^{-3} \text{ kg/m}$
C $2.0 \times 10^{-3} \text{ kg/m}$
D $4.0 \times 10^{-3} \text{ kg/m}$

Correct answer

A. $1.0 \times 10^{-3} \text{ kg/m}$

Step-by-step solution

The fundamental frequency of a stretched wire is given by:

f = \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}}

where

L

is the length,

T

is the tension, and

\mu

is the linear mass density.

Rearranging the formula to solve for

\mu

\mu = \dfrac{T}{(2Lf)^2}

Substituting the given values

L = 2.00 \text{ m}

T = 160 \text{ N}

, and

f = 100 \text{ Hz}

\mu = \dfrac{160}{(2 \times 2.00 \times 100)^2}

\mu = \dfrac{160}{(400)^2}

\mu = \dfrac{160}{160000}

\mu = 1.0 \times 10^{-3} \text{ kg/m}

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