Question

6. A transverse periodic wave on a string with a linear density of 0.200 kg/m is described by the following equation: y = 0.08 sin(469t – 28.0x), where x and y are in meters and t is in seconds. What is the tension in the string? A) 3.99 N B) 32.5 N C) 56.1 N D) 65.8 N

Answer 1

T = 56.11 N

Step-by-step explanation:

Given that,

The equation of a wave is :

y = 0.08 sin(469t – 28.0x),

where x and y are in meters and t is in seconds

The linear mass density of the wave = 0.2 kg/m

The speed of wave is given by :

`v=\sqrt{(T)/(\mu)}`

Also,

`v=(\omega)/(k)`

We have,

`k=469\ and\ \omega=28`

Put all the values,

`(\omega)/(k)=\sqrt{(T)/(\mu)}\\\\((\omega)/(k))^2=(T)/(\mu)\\\\T=((\omega)/(k))^2* \mu`

Put all the values,

`T=((469)/(28))^2* 0.2\\\\T=56.11\ N`

So, the tension in the string is 56.11 N.