MHT CET · Physics · Waves and Sound
A wave travelling along a string is described by the equation \(y=A \sin (\omega t-k x)\)
The maximum particle velocity is
- A \(A \omega\)
- B \(\omega / k\)
- C \(d \omega / d k\)
- D \(x / l\)
Answer & Solution
Correct Answer
(A) \(A \omega\)
Step-by-step Solution
Detailed explanation
Given that, the displacement of a particle is
\(
y=A \sin (\omega t-k x) ...(i)
\)
The particle velocity
\(
v_{p}=\frac{d y}{d t} ...(ii)
\)
Now, on differentiating Eq. (i) w.r.t, \(t\),
\(
\begin{aligned}
& \frac{d y}{d t}=A \cos (\omega t-k x) \cdot \omega \\
\Rightarrow \quad & \frac{d y}{d t}=A \omega \cos (\omega t-k x)
\end{aligned}
\)
From Eq. (ii)
\(
\Rightarrow \quad v_{p}=A \omega \cos (\omega t-k x)
\)
For maximum particle velocity,
\(
\cos (\omega t-k x)=1
\)
So, \(\quad v_{p}=A \omega \times 1\)
\(\Rightarrow\)
\(
v_{p}=A \omega
\)
\(
y=A \sin (\omega t-k x) ...(i)
\)
The particle velocity
\(
v_{p}=\frac{d y}{d t} ...(ii)
\)
Now, on differentiating Eq. (i) w.r.t, \(t\),
\(
\begin{aligned}
& \frac{d y}{d t}=A \cos (\omega t-k x) \cdot \omega \\
\Rightarrow \quad & \frac{d y}{d t}=A \omega \cos (\omega t-k x)
\end{aligned}
\)
From Eq. (ii)
\(
\Rightarrow \quad v_{p}=A \omega \cos (\omega t-k x)
\)
For maximum particle velocity,
\(
\cos (\omega t-k x)=1
\)
So, \(\quad v_{p}=A \omega \times 1\)
\(\Rightarrow\)
\(
v_{p}=A \omega
\)
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