
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
569.4k+ 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.
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