A wire stretched between two points with a tension $200{\text{N}}$, Whose linear density is $5 \times {10^{ - 3}}{\text{kg}}{{\text{m}}^{ - 1}}$. The wire resonates at a frequency of $170{\text{Hz}}$. The next higher frequency at which the same wire resonates is $210{\text{Hz}}$. What is the length of the wire?
Answer
561.6k+ views
Hint: Here we need to find the length of the wire and we have given frequency, tension and linear density. We need to use the relation between frequency, tension, and length for both frequencies then we substitute the known value in the equation to get the final result.
Complete step by step solution:
Let us consider that the wire vibrates at $170{\text{Hz}}$ in its ${n^{th}}$ harmonic and at $210{\text{Hz}}$ in its ${(n + 1)^{th}}$ harmonic.
Now let us use the relation between frequency, density and tension
$f = \dfrac{n}{{2L}}\sqrt {\dfrac{F}{\mu }} $
Where $f$ is the frequency, $n$ His the number of harmonic, $L$ is the length of the wire, $F$ is the tension and $\mu $ is the linear density,
Substituting the values of frequency for both the harmonics separately we get two equations
$ \Rightarrow 170 = \dfrac{n}{{2L}}\sqrt {\dfrac{F}{\mu }} - - - (1)$
$ \Rightarrow 210 = \dfrac{{(n + 1)}}{{2L}}\sqrt {\dfrac{F}{\mu }} - - - (2)$
Comparing both the equations we get,
$\dfrac{{210}}{{170}} = \dfrac{{(n + 1)}}{n}$
$ \Rightarrow 1.23n = n + 1$
$ \therefore n = \dfrac{1}{{0.23}} = 4.3$
Putting the value of $n$ and also substituting the other given values in equation 1, we get
$ 170 = \dfrac{{4.3}}{{2L}}\sqrt {\dfrac{{200}}{{5 \times {{10}^{ - 3}}}}} $
$ \Rightarrow L = \dfrac{{4.3}}{{2 \times 170}}\sqrt {40,000} $
$ \Rightarrow L = 0.0126 \times 200$
$ \therefore L = 2.5{\text{m}}$
Additional Information: For standing waves on a string the ends are fixed and the string does not move. Places, where the string is not vibrating, are called nodes. This limits the wavelengths that are possible which in turn determines the frequencies. The lowest frequency is called the fundamental or first harmonic. For a string, higher frequencies are all multiples of the fundamental and are called harmonics. The more general term overtone is used to indicate frequencies greater than the fundamental which may or may not be harmonic.
Note: We need to find the value of the number of harmonics in which the wire is in resonant condition to find the length of the wire. Linear density is the measure of a quantity of any characteristic value per unit of length.
Complete step by step solution:
Let us consider that the wire vibrates at $170{\text{Hz}}$ in its ${n^{th}}$ harmonic and at $210{\text{Hz}}$ in its ${(n + 1)^{th}}$ harmonic.
Now let us use the relation between frequency, density and tension
$f = \dfrac{n}{{2L}}\sqrt {\dfrac{F}{\mu }} $
Where $f$ is the frequency, $n$ His the number of harmonic, $L$ is the length of the wire, $F$ is the tension and $\mu $ is the linear density,
Substituting the values of frequency for both the harmonics separately we get two equations
$ \Rightarrow 170 = \dfrac{n}{{2L}}\sqrt {\dfrac{F}{\mu }} - - - (1)$
$ \Rightarrow 210 = \dfrac{{(n + 1)}}{{2L}}\sqrt {\dfrac{F}{\mu }} - - - (2)$
Comparing both the equations we get,
$\dfrac{{210}}{{170}} = \dfrac{{(n + 1)}}{n}$
$ \Rightarrow 1.23n = n + 1$
$ \therefore n = \dfrac{1}{{0.23}} = 4.3$
Putting the value of $n$ and also substituting the other given values in equation 1, we get
$ 170 = \dfrac{{4.3}}{{2L}}\sqrt {\dfrac{{200}}{{5 \times {{10}^{ - 3}}}}} $
$ \Rightarrow L = \dfrac{{4.3}}{{2 \times 170}}\sqrt {40,000} $
$ \Rightarrow L = 0.0126 \times 200$
$ \therefore L = 2.5{\text{m}}$
Additional Information: For standing waves on a string the ends are fixed and the string does not move. Places, where the string is not vibrating, are called nodes. This limits the wavelengths that are possible which in turn determines the frequencies. The lowest frequency is called the fundamental or first harmonic. For a string, higher frequencies are all multiples of the fundamental and are called harmonics. The more general term overtone is used to indicate frequencies greater than the fundamental which may or may not be harmonic.
Note: We need to find the value of the number of harmonics in which the wire is in resonant condition to find the length of the wire. Linear density is the measure of a quantity of any characteristic value per unit of length.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

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

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

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

