
The tension in a piano wire is $10\,N$. What will be the tension in a piano wire to produce a node of double the frequency?
A. $20\,N$
B. $40\,N$
C. $10\,N$
D. $120\,N$
Answer
588.6k+ views
Hint-
The relationship between frequency and tension is given as
$n = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} $
Where $n$ denotes the frequency, $T$ denotes the tension, $l$ denotes the length and $m$ denotes the mass per unit length.
Given tension in the wire,
\[{T_1} = 10\,N\]
So,
$n = \dfrac{1}{{2l}}\sqrt {\dfrac{{10}}{m}} $
We need to find the tension for double frequency. So, new frequency can be written as $2n$.
Let the corresponding tension be denoted as ${T_2}$.
Therefore,
$2n = \dfrac{1}{{2l}}\sqrt {\dfrac{{{T_2}}}{m}} $
Step by step solution:
The relationship between frequency and tension is given as
$n = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} $ …… (1)
Where $n$ denotes the frequency, $T$ denotes the tension, $l$ denotes the length and $m$ denotes the mass per unit length.
From this equation we can say that if the string is shorter then frequency will be higher if tension is higher frequency will be higher and if mass of string is less then frequency will be higher
Given tension in the wire,
\[{T_1} = 10\,N\]
Substitute this in equation (1). Then we get
$n = \dfrac{1}{{2l}}\sqrt {\dfrac{{10}}{m}} $ …… (2)
Now we need to find the tension for double frequency. That is for $2n$.
Let the corresponding tension be denoted as ${T_2}$.
Substituting in equation (1), we get
$2n = \dfrac{1}{{2l}}\sqrt {\dfrac{{{T_2}}}{m}} $ …… (3)
Dividing equation (2) by (3) we get
$\dfrac{1}{2} = \sqrt {\dfrac{{10}}{{{T_2}}}} $
Solving for ${T_2}$ we get,
$
\dfrac{1}{4} = \dfrac{{10}}{{{T_2}}} \\
{T_2} = 40\,N \\
$
Option B is correct.
Note: The answer to this question can also be found directly. We know the frequency of a vibrating string is directly proportional to square root of tension. If we increase the tension of the vibrating string to two times then the frequency will increase by $\sqrt 2 $ times. So, in order to double the frequency, we should increase the tension to four times.
The relationship between frequency and tension is given as
$n = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} $
Where $n$ denotes the frequency, $T$ denotes the tension, $l$ denotes the length and $m$ denotes the mass per unit length.
Given tension in the wire,
\[{T_1} = 10\,N\]
So,
$n = \dfrac{1}{{2l}}\sqrt {\dfrac{{10}}{m}} $
We need to find the tension for double frequency. So, new frequency can be written as $2n$.
Let the corresponding tension be denoted as ${T_2}$.
Therefore,
$2n = \dfrac{1}{{2l}}\sqrt {\dfrac{{{T_2}}}{m}} $
Step by step solution:
The relationship between frequency and tension is given as
$n = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} $ …… (1)
Where $n$ denotes the frequency, $T$ denotes the tension, $l$ denotes the length and $m$ denotes the mass per unit length.
From this equation we can say that if the string is shorter then frequency will be higher if tension is higher frequency will be higher and if mass of string is less then frequency will be higher
Given tension in the wire,
\[{T_1} = 10\,N\]
Substitute this in equation (1). Then we get
$n = \dfrac{1}{{2l}}\sqrt {\dfrac{{10}}{m}} $ …… (2)
Now we need to find the tension for double frequency. That is for $2n$.
Let the corresponding tension be denoted as ${T_2}$.
Substituting in equation (1), we get
$2n = \dfrac{1}{{2l}}\sqrt {\dfrac{{{T_2}}}{m}} $ …… (3)
Dividing equation (2) by (3) we get
$\dfrac{1}{2} = \sqrt {\dfrac{{10}}{{{T_2}}}} $
Solving for ${T_2}$ we get,
$
\dfrac{1}{4} = \dfrac{{10}}{{{T_2}}} \\
{T_2} = 40\,N \\
$
Option B is correct.
Note: The answer to this question can also be found directly. We know the frequency of a vibrating string is directly proportional to square root of tension. If we increase the tension of the vibrating string to two times then the frequency will increase by $\sqrt 2 $ times. So, in order to double the frequency, we should increase the tension to four times.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

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

Master Class 11 Economics: Engaging Questions & Answers for Success

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

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

Chemical formula of Bleaching powder is A Ca2OCl2 B class 11 chemistry CBSE

Name the part of the brain responsible for the precision class 11 biology CBSE

The growth of tendril in pea plants is due to AEffect class 11 biology CBSE

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

