Question 468: Waves

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

Accepted Answer

$$l_1$$ = 0.516 m, $$l_2$$ = 0.491 m, T = 20 N.
Mass per unit length, $$\mu $$ = 0.001 kg/m.
Frequency, $$\upsilon  = {1 \over {2l}}\sqrt {{T \over \mu }} $$
$${\upsilon _1} = {1 \over {2 	imes 0.516}}\sqrt {{{20} \over {0.001}}} $$
$${\upsilon _2} = {1 \over {2 	imes 0.491}}\sqrt {{{20} \over {0.001}}} $$
$$ 	herefore $$ Number of beats = $${\upsilon _1} - {\upsilon _2} = 7$$.

Answer

7

Answer

8

Answer

3

Answer

5

Waves

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

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