Question

. A sonometer wire 100 cm long produces a resonance with a tuning fork. When its length is decreased by 10cm, 8 beats per second are heard. Find the frequency of tuning fork. (Ans : 72 Hz.)

Answer 1

Given :

sonometer wire =l1 = 100 cm  

length is decreased by  l 2 = 100 – 10 = 90 cm  

N = n1 ~ n2 = 8/s

n =( 1/2l)√( T / m)

Since, n1 ∝  l/l1 and n2 ∝l/ 1 2  

∴  n 1/n 2 =  l2/ l1

∴ n1x l 1 = n2x l 2

100 n1 = 90 n2  

n2 = (100 /90  )n1

= 10/ 9( n1)

∵ l 1 > l 2 then n2 > n1  

n2-n1=8

[10/9 ]n1- n1=8

n1(10/ 9  – 1) = 8  

n1 = 9 × 8 = 72 Hz  

n1 = n = 72 Hz

∴the frequency of tuning fork is 72 Hz.

Answer 2

Answer:

Given :

$.$

sonometer wire =l1 = 100 cm  

length is decreased by  l 2 = 100 – 10 = 90 cm  

N = n1 ~ n2 = 8/s

n =( 1/2l)√( T / m)

Since, n1 ∝  l/l1 and n2 ∝l/ 1 2  

∴  n 1/n 2 =  l2/ l1

∴ n1x l 1 = n2x l 2

100 n1 = 90 n2  

n2 = (100 /90  )n1

= 10/ 9( n1)

∵ l 1 > l 2 then n2 > n1  

n2-n1=8

[10/9 ]n1- n1=8

n1(10/ 9  – 1) = 8  

n1 = 9 × 8 = 72 Hz  

n1 = n = 72 Hz

∴the frequency of tuning fork is 72 Hz.