Search any question & find its solution
Question:
Answered & Verified by Expert
A transverse wave given by $\mathrm{y}=2 \sin (0.01 \mathrm{x}+30 \mathrm{t})$ moves on a stretched string from one end to another end in 0.5 second. If ' $x$ ' and ' $y$ ' are in $\mathrm{cm}$ and ' $\mathrm{t}$ ' is in second, then the length of the string is
Options:
Solution:
2702 Upvotes
Verified Answer
The correct answer is:
15m
Comparing with the standard equation
$$
\mathrm{y}=\mathrm{A} \sin (\mathrm{kx}+\omega \mathrm{t})
$$
we get $\mathrm{k}=0.01 \mathrm{rad} / \mathrm{cm}$ and $\omega=30 \mathrm{rad} / \mathrm{s}$
speed of the wave, $v=\frac{\omega}{\mathrm{k}}=\frac{30}{0.01}=300 \mathrm{~cm} / \mathrm{s}=30 \mathrm{~m} / \mathrm{s}$
$\therefore$ Distance $=\mathrm{vt}=30 \times 0.5=15 \mathrm{~m}$
$$
\mathrm{y}=\mathrm{A} \sin (\mathrm{kx}+\omega \mathrm{t})
$$
we get $\mathrm{k}=0.01 \mathrm{rad} / \mathrm{cm}$ and $\omega=30 \mathrm{rad} / \mathrm{s}$
speed of the wave, $v=\frac{\omega}{\mathrm{k}}=\frac{30}{0.01}=300 \mathrm{~cm} / \mathrm{s}=30 \mathrm{~m} / \mathrm{s}$
$\therefore$ Distance $=\mathrm{vt}=30 \times 0.5=15 \mathrm{~m}$
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.