Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to 1 g/m. When both the strings vibrate simultaneously the number of beats is:

Question

Dr. Abubacker · Accepted Answer

[Solution from source] Wave speed v = √(T/μ) = √(20/10⁻³) ≈ 141.4 m/s. Fundamentals: f = v/2l gives 141.4/1.032 ≈ 137.0 Hz and 141.4/0.982 ≈ 144.0 Hz. Beat frequency = 144 − 137 = 7 per second. Answer (1).

Q. Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to 1 g/m. When both the strings vibrate simultaneously the number of beats is:

Explanation

Reviewed by Dr. Abubacker

Want to attempt the full paper?