Equation of progressive wave is given by $y = a\sin \pi \left( {\dfrac{t}{2} - \dfrac{x}{4}} \right)$, where t is in seconds and x is in meter. Then the distance through which the wave moves in 4 second is (in meter)-
A. 4m
B. 2m
C. 16m
D. 8m
Answer
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Hint- We will compare the equation given in the question to the standard equation i.e. $y = A\sin 2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\lambda }} \right)$ in order to get the values of T and $\lambda $. Then we will solve the question further.
Formula used: $y = A\sin 2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\lambda }} \right)$ and $v = f \times \lambda $.
Complete Step-by-Step solution:
The equation given to us by the question is-
$ \Rightarrow y = a\sin \pi \left( {\dfrac{t}{2} - \dfrac{x}{4}} \right)$
Let this be equation 1, we get-
$ \Rightarrow y = a\sin \pi \left( {\dfrac{t}{2} - \dfrac{x}{4}} \right)$ (equation 1)
As we know that the standard equation is-
$ \Rightarrow y = A\sin 2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\lambda }} \right)$, where $\lambda $ is wavelength and T is time period.
Let this equation be equation 2, we get-
$ \Rightarrow y = A\sin 2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\lambda }} \right)$ (equation 2)
Comparing equation 1 and equation 2, we will get-
$
\Rightarrow \lambda = 4 \\
\\
\Rightarrow T = 2 \\
$
So,
$ \Rightarrow \dfrac{1}{T} = \dfrac{1}{2} = 0.5$
As we know that the formula for frequency is $f = \dfrac{1}{T}$, substituting the value from the above equation, we get-
$ \Rightarrow f = 0.5$
Thus, F=0.5 Hz.
The formula for wave velocity is $v = f \times \lambda $
Putting the value of $f$ and $\lambda $ into the equation of velocity, we get-
$
\Rightarrow v = f \times \lambda \\
\\
\Rightarrow v = 0.5 \times 4 \\
\\
\Rightarrow v = 2cm/s \\
$
The distance through which the wave moves in 4 second is-
$ \Rightarrow v = \dfrac{d}{t}$ (where v is velocity, d is distance and t is time)
$
\Rightarrow d = t \times v \\
\\
\Rightarrow d = 2 \times 4 \\
\\
\Rightarrow d = 8m \\
$
Hence, option D is the correct option.
Note: Waves move energy starting with one spot then onto the next. In a progressive wave, the wave front travels through the medium. There are two kinds of waves which are named as- transverse and longitudinal.
Formula used: $y = A\sin 2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\lambda }} \right)$ and $v = f \times \lambda $.
Complete Step-by-Step solution:
The equation given to us by the question is-
$ \Rightarrow y = a\sin \pi \left( {\dfrac{t}{2} - \dfrac{x}{4}} \right)$
Let this be equation 1, we get-
$ \Rightarrow y = a\sin \pi \left( {\dfrac{t}{2} - \dfrac{x}{4}} \right)$ (equation 1)
As we know that the standard equation is-
$ \Rightarrow y = A\sin 2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\lambda }} \right)$, where $\lambda $ is wavelength and T is time period.
Let this equation be equation 2, we get-
$ \Rightarrow y = A\sin 2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\lambda }} \right)$ (equation 2)
Comparing equation 1 and equation 2, we will get-
$
\Rightarrow \lambda = 4 \\
\\
\Rightarrow T = 2 \\
$
So,
$ \Rightarrow \dfrac{1}{T} = \dfrac{1}{2} = 0.5$
As we know that the formula for frequency is $f = \dfrac{1}{T}$, substituting the value from the above equation, we get-
$ \Rightarrow f = 0.5$
Thus, F=0.5 Hz.
The formula for wave velocity is $v = f \times \lambda $
Putting the value of $f$ and $\lambda $ into the equation of velocity, we get-
$
\Rightarrow v = f \times \lambda \\
\\
\Rightarrow v = 0.5 \times 4 \\
\\
\Rightarrow v = 2cm/s \\
$
The distance through which the wave moves in 4 second is-
$ \Rightarrow v = \dfrac{d}{t}$ (where v is velocity, d is distance and t is time)
$
\Rightarrow d = t \times v \\
\\
\Rightarrow d = 2 \times 4 \\
\\
\Rightarrow d = 8m \\
$
Hence, option D is the correct option.
Note: Waves move energy starting with one spot then onto the next. In a progressive wave, the wave front travels through the medium. There are two kinds of waves which are named as- transverse and longitudinal.
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