Question

What is the tension in the string if the equation of travelling wave on a stretched string of linear density $5g/m$ is $y = 0.03\sin (450t - 9x)$, where distance and time are measured in SI units?
(A) $10N$
(B) $12.5N$
(C))$7.5N$
 (D) $5N$

Answer 1

Hint When we compare $y$ with ${y_0}\sin (\omega t - kx)$ we will get the values of $\omega $ and $k$.
Then, use the expression –
$v = \dfrac{\omega }{k}$ to find the value of $v$ to put it in the $v = \sqrt {\dfrac{T}{\mu }} $ (where, $\mu $ is the density) and find the value of $T$.

Complete step by step answer:
The two or more objects exert forces on each other when they are in contact with each other. If one of the objects in this is rope, string, cable or spring. Then the force is called Tension. It is denoted by $T$.
According to the question, it is given that-
Density, $\mu = 5g/m$
Equation of the travelling wave on a stretched string is $y = 0.03\sin (450t9x)$.
To get the values of $\omega $ and $k$ we will compare the above equation of $y = 0.03\sin (450t - 9x)$ with ${y_0}\sin (\omega t - kx)$
Now, we get
$\omega = 450$
$k = 9$
Now, using the formula for calculating the value of velocity $v$
$
  v = \dfrac{\omega }{k} \\
  v = \dfrac{{450}}{9} \\
  v = 50m/s \\
 $
Let the tension in the string be $T$ and density be $\mu $
Now, using the formula of
$v = \sqrt {\dfrac{T}{\mu }} $
Putting the values of velocity in the above equation
$
  \sqrt {\dfrac{T}{\mu }} = v \\
  \dfrac{T}{\mu } = {\left( {50} \right)^2} \\
 $
Making further calculation
$T = 2500 \times \mu $
Now, put the values of density in the above equation
$
  T = 2500 \times 5 \times {10^{ - 3}} \\
  T = 12.5N \\
 $
So, the tension in the string is $12.5N$

Therefore, the correct answer for this question is option (B)

Note When the tension acts on an object it is equal to the product of mass of object and gravitational force added with the product of mass and acceleration. It can be represented mathematically by-
$T = mg + ma$
The force which ropes and cables transfer over a significant distance.