
A string is clamped at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary wave is $Y=0.3Sin(0.157x)Cos(2\pi t)$. The length of the string is( All quantities are in SI units)
a)20m
b)80m
c)60m
d)40m
Answer
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Hint: In the above question it is given to us that the string is clamped at both the ends. We are also provided the equation of the stationary wave where the string is vibrating in its 4th harmonic. First we will determine the wavelength of vibration of the string by comparing the above equation to the general form. Further using the equation for length of the string vibrating in fourth harmonic we will obtain its required length.
Formula used:
$Y=ASin(kx)Cos(\omega t)$
$k=\dfrac{2\pi }{\lambda }$
$L=n\dfrac{\lambda }{2}$
Complete answer:
Let us say a string vibrates with angular velocity $\omega $, amplitude of A, wavelength $\lambda $ and the constant of propagation as k where $k=\dfrac{2\pi }{\lambda }$. Therefore the stationary wave equation for the string is in terms of its displacement of the individual fragments of the string at time ‘t’ in terms of horizontal displacement ‘x’ and vertical ‘y’ is given by,
$Y=ASin(kx)Cos(\omega t)$
In the question the wave equation for the string vibrating in the 4th harmonic is given as,
$Y=0.3Sin(0.157x)Cos(2\pi t)$
If we compare these above two equations the propagation constant in found to be equal to,
$\begin{align}
& k=0.157\text{, }\because k=\dfrac{2\pi }{\lambda } \\
& \Rightarrow \dfrac{2\pi }{\lambda }=0.157 \\
& \Rightarrow \lambda =\dfrac{2\pi }{0.157}=40m \\
\end{align}$
The length of the string clamped at both the ends for nth harmonic is given by,
$\begin{align}
& L=n\dfrac{\lambda }{2}\text{, }\because \text{n=4 } \\
& \Rightarrow L=4\dfrac{40m}{2}=2\times 40m \\
& \Rightarrow L=80m \\
\end{align}$
So, the correct answer is “Option B”.
Note:
It is to be noted that the quantities mentioned in the wave equation are in SI units and hence we would not have to make any necessary conversions to obtain the length in terms of meters. It is to be noted that the harmonics of the string basically means the mode of vibration of the string. This basically depends on where the string is plucked along its entire length.
Formula used:
$Y=ASin(kx)Cos(\omega t)$
$k=\dfrac{2\pi }{\lambda }$
$L=n\dfrac{\lambda }{2}$
Complete answer:
Let us say a string vibrates with angular velocity $\omega $, amplitude of A, wavelength $\lambda $ and the constant of propagation as k where $k=\dfrac{2\pi }{\lambda }$. Therefore the stationary wave equation for the string is in terms of its displacement of the individual fragments of the string at time ‘t’ in terms of horizontal displacement ‘x’ and vertical ‘y’ is given by,
$Y=ASin(kx)Cos(\omega t)$
In the question the wave equation for the string vibrating in the 4th harmonic is given as,
$Y=0.3Sin(0.157x)Cos(2\pi t)$
If we compare these above two equations the propagation constant in found to be equal to,
$\begin{align}
& k=0.157\text{, }\because k=\dfrac{2\pi }{\lambda } \\
& \Rightarrow \dfrac{2\pi }{\lambda }=0.157 \\
& \Rightarrow \lambda =\dfrac{2\pi }{0.157}=40m \\
\end{align}$
The length of the string clamped at both the ends for nth harmonic is given by,
$\begin{align}
& L=n\dfrac{\lambda }{2}\text{, }\because \text{n=4 } \\
& \Rightarrow L=4\dfrac{40m}{2}=2\times 40m \\
& \Rightarrow L=80m \\
\end{align}$
So, the correct answer is “Option B”.
Note:
It is to be noted that the quantities mentioned in the wave equation are in SI units and hence we would not have to make any necessary conversions to obtain the length in terms of meters. It is to be noted that the harmonics of the string basically means the mode of vibration of the string. This basically depends on where the string is plucked along its entire length.
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