Answer
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Hint: In a transverse wave, the particles of the medium vibrate along the direction perpendicular to the motion of the wave. The speed of the wave on the string is dependent on the tension in the string and mass per unit length of the given string. The exact dependence can be easily found using dimensional analysis.
Formula used: speed of the wave
$\sqrt{\dfrac{T}{\mu }}$
Complete step-by-step solution:
The speed of the transverse wave on a stretched string will be dependent on the tension in the string and on the mass per unit length of the string. We will use dimensional analysis to find the exact dependence and will get the answer to the given question. Dimensions of tension will be the same as Force and dimensions of μ i.e. mass per unit length will be of mass divided by length.
$\begin{align}
& v\propto {{T}^{a}}{{\mu }^{b}} \\
& [L{{T}^{-1}}]={{[ML{{T}^{-2}}]}^{a}}{{[M{{L}^{-1}}]}^{b}}=[{{M}^{a+b}}{{L}^{a-b}}{{T}^{-2a}}] \\
\end{align}$
We get the following relations from the expression above
$a+b = 0$
$a-b = 1$
$-2a = -1$
When we solve them we get a = $\dfrac{1}{2}$ and b = \[-\dfrac{1}{2}\]
So, it can be seen that v is proportional to the square root of the tension. Hence, the correct option is B, i.e. directly proportional to the square root of the tension.
Note: We can also find the exact relation using other rigorous mathematical methods, but that will be very tedious and time-consuming so it is advised to stick to dimensional analysis as it is easy and also gets us the desired result and the correct solution with minimal effort.
Formula used: speed of the wave
$\sqrt{\dfrac{T}{\mu }}$
Complete step-by-step solution:
The speed of the transverse wave on a stretched string will be dependent on the tension in the string and on the mass per unit length of the string. We will use dimensional analysis to find the exact dependence and will get the answer to the given question. Dimensions of tension will be the same as Force and dimensions of μ i.e. mass per unit length will be of mass divided by length.
$\begin{align}
& v\propto {{T}^{a}}{{\mu }^{b}} \\
& [L{{T}^{-1}}]={{[ML{{T}^{-2}}]}^{a}}{{[M{{L}^{-1}}]}^{b}}=[{{M}^{a+b}}{{L}^{a-b}}{{T}^{-2a}}] \\
\end{align}$
We get the following relations from the expression above
$a+b = 0$
$a-b = 1$
$-2a = -1$
When we solve them we get a = $\dfrac{1}{2}$ and b = \[-\dfrac{1}{2}\]
So, it can be seen that v is proportional to the square root of the tension. Hence, the correct option is B, i.e. directly proportional to the square root of the tension.
Note: We can also find the exact relation using other rigorous mathematical methods, but that will be very tedious and time-consuming so it is advised to stick to dimensional analysis as it is easy and also gets us the desired result and the correct solution with minimal effort.
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