The speed of a transverse wave on a stretched string is:
A). directly proportional to the tension in the string.
B). directly proportional to the square root of the tension.
C). inversely proportional to tension.
D). inversely proportional to the square root of the tension.
Answer
613.2k+ views
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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

