JEE PYQ: Waves And Sound Question 43
Question 43 - 2019 (11 Jan Shift 1)
Equation of travelling wave on a stretched string of linear density 5 g/m is $y = 0.03\sin(450t - 9x)$ where distance and time are measured in SI units. The tension in the string is:
(1) 10 N
(2) 7.5 N
(3) 12.5 N
(4) 5 N
Show Answer
Answer: (3)
Solution
We have given,
$y = 0.03\sin(450t - 9x)$
Comparing it with standard equation of wave, we get
$\omega = 450$, $k = 9$
$\therefore v = \frac{\omega}{k} = \frac{450}{9} = 50$ m/s
Velocity of travelling wave on a stretched string is given by
$v = \sqrt{\frac{T}{\mu}} \Rightarrow \frac{T}{\mu} = 2500$
$\mu =$ linear mass density $= 5 \times 10^{-3}$
$\Rightarrow T = 2500 \times 5 \times 10^{-3}$
$\Rightarrow T = 12.5$ N
BETA
