JEE MAIN - Physics (2023 - 15th April Morning Shift - No. 20)

The fundamental frequency of vibration of a string stretched between two rigid support is $50 \mathrm{~Hz}$. The mass of the string is $18 \mathrm{~g}$ and its linear mass density is $20 \mathrm{~g} / \mathrm{m}$. The speed of the transverse waves so produced in the string is ___________ $\mathrm{ms}^{-1}$
Answer
90

Explanation

To find the speed of the transverse waves produced in the string, we can use the formula for the fundamental frequency of a vibrating string:

$f = \frac{1}{2L} \cdot v$

where $f$ is the fundamental frequency, $L$ is the length of the string, and $v$ is the speed of the transverse waves.

First, we are given the mass of the string ($m = 18g$) and the linear mass density ($\mu = 20g/m$). We can find the length of the string by dividing the mass by the linear mass density:

$L = \frac{m}{\mu} = \frac{18g}{20g/m} = 0.9m$

Now we can plug in the values for the fundamental frequency ($f = 50Hz$) and the length of the string ($L = 0.9m$) into the formula:

$50Hz = \frac{1}{2(0.9m)} \cdot v$

To isolate $v$, we multiply both sides by $2(0.9m)$:

$v = 50Hz \cdot 2(0.9m) = 90 \mathrm{ms}^{-1}$

The speed of the transverse waves produced in the string is $90 ~\mathrm{ms}^{-1}$.

Alternate Method:

To find the speed of the transverse waves produced in the string, we can use the formula for the fundamental frequency of a vibrating string:

$f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$

where $f_1$ is the fundamental frequency, $L$ is the length of the string, $T$ is the tension in the string, and $\mu$ is the linear mass density of the string.

We're given that the fundamental frequency $f_1 = 50 ,\text{Hz}$, the mass of the string $m = 18 ,\text{g}$, and the linear mass density $\mu = 20 ,\text{g/m}$. To find the speed of the transverse waves, we need to find the tension $T$ and the length $L$ of the string.

First, let's find the length $L$ of the string using the mass and linear mass density:

$L = \frac{m}{\mu} = \frac{18 ,\text{g}}{20 ,\text{g/m}} = 0.9 ~\text{m}$

Now, we can rearrange the formula for the fundamental frequency to solve for the tension $T$:

$T = \mu \left(\frac{2Lf_1}{1}\right)^2$

Substitute the known values:

$T = 20 ,\text{g/m} \cdot \left(\frac{2 \cdot 0.9 ~\text{m} \cdot 50 ~\text{Hz}}{1}\right)^2$

$T = 20 ,\text{g/m} \cdot (90 ~\text{m/s})^2$

$T = 20 ,\text{g/m} \cdot 8100 ~\text{m}^2/\text{s}^2$

$T = 162000 ,\text{g m}/\text{s}^2$

Now, we can find the speed of the transverse waves $v$ using the formula:

$v = \sqrt{\frac{T}{\mu}}$

Substitute the known values:

$v = \sqrt{\frac{162000}{20}}$

$v = \sqrt{8100} = 90~ \text{m/s}$

The speed of the transverse waves produced in the string is $90 ~\text{m/s}$.

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