
In a transverse wave the distance between a crest and neighboring trough at the same instant is 2.0 cm and the distance between a crest and trough at the same place is 1.0 cm. The next crest appears at the same place after a time interval of 0.4s. The maximum speed of vibrating particles in the medium is
A. $\dfrac{{5\pi }}{2}cm/s$
B. $2\pi cm/s$
C. $\dfrac{\pi }{2}cm/s$
D. $\dfrac{{3\pi }}{2}cm/s$
Answer
497.4k+ views
Hint – The distance between two consecutive crests or troughs or between a crest and neighboring trough is the wavelength of the wave and the number of times a crest or trough appears in the wave is called the time period of the wave.
Formula used – 1) Maximum velocity of vibrating particle=$A\omega $
2) Angular velocity, $\omega $= $\dfrac{{2\pi }}{T}$
Complete step-by-step answer:
Given,
Distance between crest and neighboring trough =2.0cm
Distance between crest and trough at the same place=1.0cm
Time interval=0.4s
Time period of wave =0.4s
Angular velocity, $\omega $= $\dfrac{{2\pi }}{T} = \dfrac{{2\pi }}{{(0.4)}} = 5\pi $
Amplitude=$
y = A\sin (\omega t + kx) \\
v = A\omega = \dfrac{1}{2} \times (5\pi ) \\
= \dfrac{{5\pi }}{2}cm/s \\
$ $\dfrac{1}{2} = 0.5cm$
The transverse wave equation is be given by,
$y = A\sin (\omega t + kx)$
Maximum velocity= $v = A\omega = \dfrac{1}{2} \times (5\pi )$
$ = \dfrac{{5\pi }}{2}cm/s$
Hence, the correct option is option A.
Note – In these types of questions, we need to know the basic concepts of wave as between two consecutive crests or trough or between a crest and neighboring trough is the wavelength of the wave and the number of time a crest or trough appears in the wave is called the time period of the wave. The sine wave is important in physics because it preserves its waveform as it is connected to another sine wave of the same frequency and arbitrary phase and magnitude. It's the only periodic waveform that has this property. This property has an influence on Fourier 's analysis and makes it acoustically distinct.
Formula used – 1) Maximum velocity of vibrating particle=$A\omega $
2) Angular velocity, $\omega $= $\dfrac{{2\pi }}{T}$
Complete step-by-step answer:
Given,
Distance between crest and neighboring trough =2.0cm
Distance between crest and trough at the same place=1.0cm
Time interval=0.4s
Time period of wave =0.4s
Angular velocity, $\omega $= $\dfrac{{2\pi }}{T} = \dfrac{{2\pi }}{{(0.4)}} = 5\pi $
Amplitude=$
y = A\sin (\omega t + kx) \\
v = A\omega = \dfrac{1}{2} \times (5\pi ) \\
= \dfrac{{5\pi }}{2}cm/s \\
$ $\dfrac{1}{2} = 0.5cm$
The transverse wave equation is be given by,
$y = A\sin (\omega t + kx)$
Maximum velocity= $v = A\omega = \dfrac{1}{2} \times (5\pi )$
$ = \dfrac{{5\pi }}{2}cm/s$
Hence, the correct option is option A.
Note – In these types of questions, we need to know the basic concepts of wave as between two consecutive crests or trough or between a crest and neighboring trough is the wavelength of the wave and the number of time a crest or trough appears in the wave is called the time period of the wave. The sine wave is important in physics because it preserves its waveform as it is connected to another sine wave of the same frequency and arbitrary phase and magnitude. It's the only periodic waveform that has this property. This property has an influence on Fourier 's analysis and makes it acoustically distinct.
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