JEE PYQ: Waves And Sound Question 40
Question 40 - 2019 (10 Jan Shift 1)
A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to:
(1) 10.0 cm
(2) 33.3 cm
(3) 16.6 cm
(4) 20.0 cm
Show Answer
Answer: (4)
Solution
Velocity of wave on string:
$V = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{8}{5} \times 1000} = 40$ m/s
Wavelength of wave $\lambda = \frac{v}{n} = \frac{40}{100}$ m
Separation b/w successive nodes,
$\frac{\lambda}{2} = \frac{40}{2 \times 100} = \frac{20}{100}$ m $= 20$ cm
BETA
