JEE PYQ: Waves And Sound Question 26
Question 26 - 2019 (09 Apr Shift 1)
A string is clamped at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary wave is $Y = 0.3\sin(0.157x)\cos(200\pi t)$. The length of the string is: (All quantities are in SI units.)
(1) 20 m
(2) 80 m
(3) 40 m
(4) 60 m
Show Answer
Answer: (2)
Solution
Given, $y = 0.3\sin(0.157x)\cos(200\pi t)$
So, $k = 0.157$ and $\omega = 200\pi = 2\pi f$
$\therefore f = 100$ Hz and $v = \frac{\omega}{k} = \frac{200\pi}{0.157} = 4000$ m/s
Now, using $f = \frac{nv}{2l} = \frac{4v}{2l}$ [here $n = 4$]
$\therefore l = \frac{2v}{f} = \frac{2 \times 4000}{100} = 80$ m
BETA
