Question

Each of the two string of length 51.6cm and 49.1cm are

Answer 1

Answer:

Each of the two strings of length 51.6cm and 49.1 cm are tensioned separately by 20N force. Mass per unit length of both the strings is same and equal to 1g/m. When both the strings vibrate simultaneously the no of beats is. 1- 3 2- 5.

Answer 2

The number of beats will be the difference of frequencies of the two strings.

frequency of first string

f

1

=

21

1

1

m

T

=

2×51.6×10

−2

1

10

−3

20

similarly, frequency of second string

=

2×49.1×10

−2

1

10

−3

20

Number of beats = f

2

−f

1

= 144 -137

=7 beats