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 gm ⁻¹ . When both the strings vibrate simultaneously the number of beats is

Step-by-step solution

Key Idea The number of beats will be the difference of frequencies of the two strings. Frequency of first string f₁= 1 2 l₁ T m aligned & = 1 2 51.6 10⁻² 20 10⁻³ & =137.03 ~Hz aligned Similarly, frequency of second string aligned & = 1 2 49.1 10⁻² 20 10⁻³ & =144.01 aligned Number of beats = f ₂- f ₁=144-137=7 beats