Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A wave in a string has an amplitude of $2 \mathrm{~cm}$. The wave travels in the +ve direction of $x$ axis with a speed of $128 \mathrm{~ms}^{-1}$ and it is noted that 5 complete waves fit in $4 \mathrm{~m}$ length of the string. The equation describing the wave is
PhysicsWaves and SoundNEETNEET 2009 (Screening)
Options:
  • A $y=(0.02) \mathrm{m} \sin (7.85 x+1005 \mathrm{t})$
  • B $\mathrm{y}=(0.02) \mathrm{m} \sin (15.7 \mathrm{x}-2010 \mathrm{t})$
  • C $y=(0.02) \mathrm{m} \sin (15.7 \mathrm{x}+2010 \mathrm{t})$
  • D $y=(0.02) m \sin (7.85 x-1005 \mathrm{t})$
Solution:
1411 Upvotes Verified Answer
The correct answer is: $y=(0.02) m \sin (7.85 x-1005 \mathrm{t})$
Key Idea Find the parameters and put in the general wave equation.
Here,
$$
\begin{aligned}
A & =2 \mathrm{~cm} \\
\text { direction } & =+ \text { ve direction } \\
\mathrm{v} & =128 \mathrm{~ms}^{-1}
\end{aligned}
$$
and
$$
5 \lambda=4
$$
Now,
$$
\mathrm{k}=\frac{2 \pi}{\lambda}=\frac{2 \pi \times 5}{4}=7.85
$$
and
$$
\mathrm{v}=\frac{\omega}{\mathrm{k}}=128 \mathrm{~ms}^{-1}
$$
$$
\begin{aligned}
\Rightarrow \quad \omega & =\mathrm{v} \times \mathrm{k}=128 \times 7.85 \\
& =1005
\end{aligned}
$$


$$
\begin{array}{lll}
\text { As, } & y & =\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t}) \\
\therefore & & \mathrm{y}=2 \sin (7.85 \mathrm{x}-1005 \mathrm{t}) \\
& & =(0.02) \mathrm{m} \sin (7.85 \mathrm{x}-1005 \mathrm{t})
\end{array}
$$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.