KCET · Physics · Waves and Sound
The equation of a transverse wave is given by \(y=0.05 \sin \pi(2 t-0.02 x)\), where \(x, y\) are in metre and \(t\) is in second. The minimum distance of separation between two particles which are in phase and the wave velocity are respectively
- A \(50 \mathrm{~m}, 50 \mathrm{~ms}^{-1}\)
- B \(100 \mathrm{~m}, 100 \mathrm{~ms}^{-1}\)
- C \(50 \mathrm{~m}, 100 \mathrm{~ms}^{-1}\)
- D \(100 \mathrm{~m}, 50 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(B) \(100 \mathrm{~m}, 100 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, \(y=0.05 \sin \pi(2 t-0.02 x)\)
or \(\quad y=0.05 \sin (2 \pi t-0.02 \pi x)\)
On comparing this equation with standard equation
\(y=a \sin (\omega t-k x)\)
where \(\omega=\frac{2 \pi}{T}\) and \(K=\frac{2 \pi}{\lambda}\), we have
\(\frac{2 \pi}{\lambda}=0.02 \pi \Rightarrow \lambda=100 \mathrm{~m}\)
Also, \(\quad \omega=\frac{2 \pi}{T}=2 \pi\)
or \(\quad \frac{1}{T}=1 \Rightarrow v=1\)
So, \(v=r \lambda=1 \times 100=100 \mathrm{~m} / \mathrm{s}\)
or \(\quad y=0.05 \sin (2 \pi t-0.02 \pi x)\)
On comparing this equation with standard equation
\(y=a \sin (\omega t-k x)\)
where \(\omega=\frac{2 \pi}{T}\) and \(K=\frac{2 \pi}{\lambda}\), we have
\(\frac{2 \pi}{\lambda}=0.02 \pi \Rightarrow \lambda=100 \mathrm{~m}\)
Also, \(\quad \omega=\frac{2 \pi}{T}=2 \pi\)
or \(\quad \frac{1}{T}=1 \Rightarrow v=1\)
So, \(v=r \lambda=1 \times 100=100 \mathrm{~m} / \mathrm{s}\)
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