Question

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 :

Answer 1

Answer:

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