Question

Each of the two string of length 51.6 cm and 49.1 cm attention separately by 20 newton force mass per unit length of the string is same and equal to 1 gram per metre

Answer 1

Given Each of the two string of length 51.6 cm and 49.1 cm attention separately by 20 newton force mass per unit length of the string is same and equal to 1 gram per metre. The number of beats will be

We know that frequency of the string f = 1/2l √T/m

So for first string F = 1 /2 x 51.6 x 10⁻² √20/10⁻³

                    F = 100 / 103.2 √20,000

                 F = 137.03

Frequency of second string f = 1/2 x 49.1 x 10⁻²√20000

                                             f = 144.01

So difference in frequency of two strings = f - F = 144.01 - 137.03 = 7.02 or 7 beats

Answer 2

Answer:

7

Explanation: