
A sonometer wire is to be divided into three segments having fundamental frequencies in the ratio 1 : 2 : 3. What should be the ratio of lengths?
A. 4 : 2 : 1
B. 4 : 3 : 1
C. 6 : 3 : 2
D. 3 : 2 : 1
Answer
573.9k+ views
Hint: To solve this question, we must know the formula for the fundamental frequency of wire with both the ends fixed. The frequency depends on the length, tension in the wire and mass per unit of the wire. However, the tension and mass per unit length are the same for all three segments.
Formula used:
$f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu }}$
Complete answer:
A sonometer wire is a stretched wire. When the wire is vibrated at a point, it undergoes oscillations. These oscillations are called harmonics.
The first harmonic is when the wave has two nodes and only antinodes. Nodes are that points on the wires that have minimum amplitudes and antinodes are those points that have the maximum amplitudes.
The first harmonic is also called a fundamental frequency. The value of the fundamental frequency of a sonometer wire is given as $f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu }}$.
Here, l is the length of the wire that under the oscillation, T is the tension in the wire and $\mu $ is the mass per unit length of the wire.
Since we are making the three segments of the same wire, the tension T and the mass per unit length $\mu $ will be the same for all the three segments.
Let the length of the segments with frequencies ${{f}_{1}}$, ${{f}_{2}}$ and ${{f}_{3}}$ be ${{l}_{1}}$, ${{l}_{2}}$ and ${{l}_{3}}$.
It is given that ${{f}_{1}}$ : ${{f}_{2}}$ : ${{f}_{3}}$ = 1 : 2 : 3.
Therefore, $\dfrac{1}{{{l}_{1}}}:\dfrac{1}{{{l}_{2}}}:\dfrac{1}{{{l}_{3}}}=1:2:3$.
$\Rightarrow {{l}_{1}}:{{l}_{2}}:{{l}_{3}}=1:\dfrac{1}{2}:\dfrac{1}{3}$
$\Rightarrow {{l}_{1}}:{{l}_{2}}:{{l}_{3}}=6:3:2$.
Hence, the correct option is C.
Note:
We can see that the segment having the longest length is having the lowest frequency and the segment having the shortest length has the highest frequency.
Therefore, this means that the frequency is inversely proportional to the length of the segments.
Formula used:
$f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu }}$
Complete answer:
A sonometer wire is a stretched wire. When the wire is vibrated at a point, it undergoes oscillations. These oscillations are called harmonics.
The first harmonic is when the wave has two nodes and only antinodes. Nodes are that points on the wires that have minimum amplitudes and antinodes are those points that have the maximum amplitudes.
The first harmonic is also called a fundamental frequency. The value of the fundamental frequency of a sonometer wire is given as $f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu }}$.
Here, l is the length of the wire that under the oscillation, T is the tension in the wire and $\mu $ is the mass per unit length of the wire.
Since we are making the three segments of the same wire, the tension T and the mass per unit length $\mu $ will be the same for all the three segments.
Let the length of the segments with frequencies ${{f}_{1}}$, ${{f}_{2}}$ and ${{f}_{3}}$ be ${{l}_{1}}$, ${{l}_{2}}$ and ${{l}_{3}}$.
It is given that ${{f}_{1}}$ : ${{f}_{2}}$ : ${{f}_{3}}$ = 1 : 2 : 3.
Therefore, $\dfrac{1}{{{l}_{1}}}:\dfrac{1}{{{l}_{2}}}:\dfrac{1}{{{l}_{3}}}=1:2:3$.
$\Rightarrow {{l}_{1}}:{{l}_{2}}:{{l}_{3}}=1:\dfrac{1}{2}:\dfrac{1}{3}$
$\Rightarrow {{l}_{1}}:{{l}_{2}}:{{l}_{3}}=6:3:2$.
Hence, the correct option is C.
Note:
We can see that the segment having the longest length is having the lowest frequency and the segment having the shortest length has the highest frequency.
Therefore, this means that the frequency is inversely proportional to the length of the segments.
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