An aluminium rod having a length \[100cm\] is clamped at its middle point and set into longitude vibrations let the rod vibrate in its fundamental mode. The density of aluminium is $2600kg/m^{3}$ and its Young’s modulus is $7.8\times 10^{10}N/m^{2}$, the frequency of the sound produced is:
\[\begin{align}
& A.1250Hz \\
& B.2740Hz \\
& C.2350Hz \\
& D.1685Hz \\
\end{align}\]
Answer
600.9k+ views
Hint: A mode is observed in an oscillating system, where the waves are sinusoidal in nature, with a fixed phase difference. Generally the motion of the system is along the superposition of the modes. Also the oscillating system has a fixed or constant frequency.
Formula used:
$f=\dfrac{v}{\lambda}$
Complete step-by-step answer:
Given that the length of the rod is $l=100cm$ ,and is clamped at the centre. Also given that the density of aluminium rod is $\rho=2600kg/m^{3}$ and the Young’s modulus of the given rod is $Y=7.8\times 10^{10}N/m^{2}$.
We know that antinodes are formed at the middle of the rod. Since the rod is clamped at the centre, then clearly the ends of the rod indicate the nodes. Clearly, the rod is clamped at the antinode of the rod and also there are no other nodes or antinodes in the rod.
Then we can calculate the velocity $v$ at the antinode as $v=\sqrt{\dfrac{Y}{\rho}}=\sqrt{\dfrac{7.8\times 10^{10}}{2600}}=5477m/s$
Since the vibration is observed at the fundamental mode, we can say that $\dfrac{\lambda}{2}=l$
On rearranging, we get $\lambda=2l$
On substitution of the value for $l$ we get, $\lambda=2\times 100=200cm=2m$
We know that the frequency is given as $f=\dfrac{v}{\lambda}$
Since we found the value of $\lambda$ and $v$, substituting, we get $f=\dfrac{5477}{2}=2738 Hz\approx 2740 Hz$
Hence we get the frequency of the sound produced due to the vibration of the aluminium rod is given as \[2740Hz\]
Hence the answer is \[B.2740Hz\]
So, the correct answer is “Option B”.
Note: Nodes are different from modes. Nodes are a special condition of modes, where the displacement of the system is zero. In one-dimension, the system generally refers to a point, and in two-dimensions they refer to a line. Here, the frequency of the rod is the frequency of the sound produced.
Formula used:
$f=\dfrac{v}{\lambda}$
Complete step-by-step answer:
Given that the length of the rod is $l=100cm$ ,and is clamped at the centre. Also given that the density of aluminium rod is $\rho=2600kg/m^{3}$ and the Young’s modulus of the given rod is $Y=7.8\times 10^{10}N/m^{2}$.
We know that antinodes are formed at the middle of the rod. Since the rod is clamped at the centre, then clearly the ends of the rod indicate the nodes. Clearly, the rod is clamped at the antinode of the rod and also there are no other nodes or antinodes in the rod.
Then we can calculate the velocity $v$ at the antinode as $v=\sqrt{\dfrac{Y}{\rho}}=\sqrt{\dfrac{7.8\times 10^{10}}{2600}}=5477m/s$
Since the vibration is observed at the fundamental mode, we can say that $\dfrac{\lambda}{2}=l$
On rearranging, we get $\lambda=2l$
On substitution of the value for $l$ we get, $\lambda=2\times 100=200cm=2m$
We know that the frequency is given as $f=\dfrac{v}{\lambda}$
Since we found the value of $\lambda$ and $v$, substituting, we get $f=\dfrac{5477}{2}=2738 Hz\approx 2740 Hz$
Hence we get the frequency of the sound produced due to the vibration of the aluminium rod is given as \[2740Hz\]
Hence the answer is \[B.2740Hz\]
So, the correct answer is “Option B”.
Note: Nodes are different from modes. Nodes are a special condition of modes, where the displacement of the system is zero. In one-dimension, the system generally refers to a point, and in two-dimensions they refer to a line. Here, the frequency of the rod is the frequency of the sound produced.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

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

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

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

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

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

