
Each of the two strings of length $51.6cm$ and $49.1cm$ are tensioned separately by $20N$ force. Mass per unit length of both the strings is the same and equal to $1g{m^{ - 1}}$ . When both strings vibrate simultaneously the number of beats is:
A) $7$
B) $8$
C) $3$
D) $5$
Answer
232.8k+ views
Hint: In the above question we are given two strings of different lengths. Now, we know that the number of beats is given by the difference between the frequencies of the two strings. Now, by calculating the frequencies of the strings using the formula and then subtracting them, will give us the value of the beats.
Formula used: We will use the formula for frequency of a vibrating string.
Which is $f = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} $ , where, $l$ is the length of the string, $T$ is the tension in the string and $m$ is the mass of the string.
Complete step by step solution:
In the above question we are given that two strings of different lengths are tensioned by a force.
Now, to calculate the number of beats when both the strings are vibrating, we will find the frequencies of the strings and then subtract them.
Now, we know that frequency of a vibrating string is $f = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} $, where, $l$ is the length of the string, $T$ is the tension in the string and $m$ is the mass of the string.
So, the frequency of the first string is ${f_1} = \dfrac{1}{{2{l_1}}}\sqrt {\dfrac{T}{m}} $ and second string is ${f_2} = \dfrac{1}{{2{l_2}}}\sqrt {\dfrac{T}{m}} $
Now, we know that number of beats is ${n_b} = {f_2} - {f_1}$
So, ${n_b} = \dfrac{1}{{2{l_1}}}\sqrt {\dfrac{T}{m}} + \dfrac{1}{{2{l_2}}}\sqrt {\dfrac{T}{m}} $
Now, substituting the length, mass and tension in the above equation,
$
{n_b} = \dfrac{1}{{2{l_1}}}\sqrt {\dfrac{T}{m}} + \dfrac{1}{{2{l_2}}}\sqrt {\dfrac{T}{m}} \\
{n_b} = \dfrac{1}{2}\sqrt {\dfrac{{20}}{{{{10}^{ - 3}}}}} \left( {\dfrac{1}{{49.1 \times {{10}^{ - 2}}}} - \dfrac{1}{{51.6 \times {{10}^{ - 2}}}}} \right) \\
{n_b} = 144 - 137 \\
{n_b} = 7 \\
$
Hence, the number of beats is 7.
Hence, the correct option is (A).
Note: we know when string is tensioned with a force, it vibrates with a frequency. Now, we also know that beat is produced when the two notes of the different frequencies interfere. Now, by subtracting the frequencies, we can calculate the number of beats.
Formula used: We will use the formula for frequency of a vibrating string.
Which is $f = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} $ , where, $l$ is the length of the string, $T$ is the tension in the string and $m$ is the mass of the string.
Complete step by step solution:
In the above question we are given that two strings of different lengths are tensioned by a force.
Now, to calculate the number of beats when both the strings are vibrating, we will find the frequencies of the strings and then subtract them.
Now, we know that frequency of a vibrating string is $f = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} $, where, $l$ is the length of the string, $T$ is the tension in the string and $m$ is the mass of the string.
So, the frequency of the first string is ${f_1} = \dfrac{1}{{2{l_1}}}\sqrt {\dfrac{T}{m}} $ and second string is ${f_2} = \dfrac{1}{{2{l_2}}}\sqrt {\dfrac{T}{m}} $
Now, we know that number of beats is ${n_b} = {f_2} - {f_1}$
So, ${n_b} = \dfrac{1}{{2{l_1}}}\sqrt {\dfrac{T}{m}} + \dfrac{1}{{2{l_2}}}\sqrt {\dfrac{T}{m}} $
Now, substituting the length, mass and tension in the above equation,
$
{n_b} = \dfrac{1}{{2{l_1}}}\sqrt {\dfrac{T}{m}} + \dfrac{1}{{2{l_2}}}\sqrt {\dfrac{T}{m}} \\
{n_b} = \dfrac{1}{2}\sqrt {\dfrac{{20}}{{{{10}^{ - 3}}}}} \left( {\dfrac{1}{{49.1 \times {{10}^{ - 2}}}} - \dfrac{1}{{51.6 \times {{10}^{ - 2}}}}} \right) \\
{n_b} = 144 - 137 \\
{n_b} = 7 \\
$
Hence, the number of beats is 7.
Hence, the correct option is (A).
Note: we know when string is tensioned with a force, it vibrates with a frequency. Now, we also know that beat is produced when the two notes of the different frequencies interfere. Now, by subtracting the frequencies, we can calculate the number of beats.
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