
A stretched string of length l, fixed at both ends, can sustain stationary waves of wavelength λ, correctly given by
$\begin{align}
& \left( A \right)\lambda =\dfrac{{{l}^{2}}}{2p} \\
& \left( B \right)\lambda =\dfrac{{{p}^{2}}}{2l} \\
& \left( C \right)\lambda =2lp \\
& \left( D \right)\lambda =\dfrac{2l}{p} \\
\end{align}$
Answer
511.8k+ views
Hint: If the ends of the string are fixed then only vibrational modes are allowed. When two progressive waves of same amplitude and wavelength travelling along a line in opposite directions superimpose on one another, stationary waves are formed.
Complete step by step answer:
As given above that the string has a length $l$ and both ends are fixed then only vibrational modes are allowed where an integer number of half-wavelengths equals the string length. Now the fundamental wavelength in the wire is:
$\lambda =\dfrac{2l}{p}$ where, \[p=1,2,3..\]
So, the correct option for this is option (D).
Additional Information:
The waveform remains stationary. Nodes and antinodes are formed alternately.
Where displacement is zero, the points are called nodes, and thus antinodes are considered the points where the displacement is maximal. Changes in pressure are highest at nodes, and small at antinodes. During a standing stream, nodes and antinodes. Except those at the nodes, all the particles perform simple harmonic motions of the same duration. In musical instruments like sitar, violin, etc. sound is produced thanks to the vibrations of the stretched strings. Here, we shall discuss the various modes of vibrations of a string which is rigidly fixed at both ends. When a string under tension is about into vibration, a transverse progressive wave moves towards the top of the wire and gets reflected. Thus, stationary waves are formed.
Note:
Distance between any two consecutive nodes or antinodes is adequate to $\dfrac{\lambda }{2}$, whereas the space between a node and its adjacent antinode is adequate to $\dfrac{\lambda }{4}$. Vibration mode of string depends upon the fixed point of the string.
Complete step by step answer:
As given above that the string has a length $l$ and both ends are fixed then only vibrational modes are allowed where an integer number of half-wavelengths equals the string length. Now the fundamental wavelength in the wire is:
$\lambda =\dfrac{2l}{p}$ where, \[p=1,2,3..\]
So, the correct option for this is option (D).
Additional Information:
The waveform remains stationary. Nodes and antinodes are formed alternately.
Where displacement is zero, the points are called nodes, and thus antinodes are considered the points where the displacement is maximal. Changes in pressure are highest at nodes, and small at antinodes. During a standing stream, nodes and antinodes. Except those at the nodes, all the particles perform simple harmonic motions of the same duration. In musical instruments like sitar, violin, etc. sound is produced thanks to the vibrations of the stretched strings. Here, we shall discuss the various modes of vibrations of a string which is rigidly fixed at both ends. When a string under tension is about into vibration, a transverse progressive wave moves towards the top of the wire and gets reflected. Thus, stationary waves are formed.
Note:
Distance between any two consecutive nodes or antinodes is adequate to $\dfrac{\lambda }{2}$, whereas the space between a node and its adjacent antinode is adequate to $\dfrac{\lambda }{4}$. Vibration mode of string depends upon the fixed point of the string.
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