Question

A transverse wave on a stretched string of linear density 3 10^-4 is represented by y=0.2sin(15 x+60 t) the tension in the string is

Answer 1

Answer:

Given:

Wave Equation of a travelling wave as follows ;

$y = 0.2 \sin(15x + 60t)$

Linear density of the string is :

$\mu = 3 \times {10}^{ - 4} kg {m}^{ - 1}$

To find:

Tension in the string

Concept :

Whenever a transverse pulse flows through a string , we can explain it's movement by perpendicular Oscillation of medium particles.

The wave Velocity can be related to Tension and linear mass density in the following manner :

$\boxed{ \huge{ \red{ \bold{v = \sqrt{ \dfrac{T}{\mu} } }}}}$

Calculation:

From the wave Equation , Velocity of the wave can be calculated as follows :

$v = \dfrac{ \omega}{k}$

$= > v = \dfrac{60}{15}$

$= > v = 4 \: m {s}^{ - 1}$

Now putting all the values in the equation :

$\blue{ \sf{ = > 4 = \sqrt{ \dfrac{T}{ \mu} } }}$

$\blue{ \sf{ = > 4 = \sqrt{ \dfrac{T}{3 \times {10}^{ - 4} } } }}$

$\blue{ \sf{ = > 16 = \dfrac{T}{3 \times {10}^{ - 4} } }}$

$\blue{ \sf{ = > T = 48 \times {10}^{ - 4} N }}$

So final answer :

$\green{ \sf{ \bold{ \huge{T = 48 \times {10}^{ - 4} N }}}}$

Answer 2

$\huge\fcolorbox{red}{pink}{Answer}$

Given :

$\implies \rm \: transverse \: wave \: equation \\ \\ \: \: \: \: \: \: \: \: \boxed{ \rm{ \red{y = 0.2 \sin(15x + 60t) }}} \\ \\ \implies \rm \: linear \: mass \: density \: of \: string \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \rm{ \red{ \mu = 3 \times {10}^{ - 4} \: \frac{kg}{m}}}}$

To Find :

Tension in the string ??

Formula :

Wave velocity is given by...

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \dag \: \underline{ \boxed{ \bold{ \rm{ \pink{V = \frac{ \omega}{k}}}}}} \: \dag$

Tension in the string in terms of wave velocity and linear mass density is given by...

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \dag \: \underline{ \boxed{ \bold{ \rm{ \blue{V = \sqrt{ \frac{T}{ \mu}}}}}}} \: \dag$

Calculation :

$\therefore \rm \: \frac{ \omega}{k} = \sqrt{ \frac{T}{ \mu} } \\ \\ \therefore \rm \: \frac{60}{15} = \sqrt{ \frac{T}{3 \times {10}^{ - 4} } } \\ \\ \therefore \rm \: T = 16 \times 3 \times {10}^{ - 4} \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \orange{T =4.8 \times {10}^{ - 3} \: N }}}}} \: \purple{ \clubsuit}$