Two wires of the same material and having the same radius have their fundamental frequencies in the ratio 1:2 and under tension in the ratio 1:8, Compare ratio of their lengths. (Ans : 0.7072:1)
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Answered by
15
Given ;
n 1/n 2 = 1/ 2
T 1/T 2 = 1/ 8
n =( 1/2l)√( T/2m)
n1=(1/2l1) √ (T1/m)
l1=1/2n1 √ (T1/m) ----1
for second case
n2=(/2) l2√(T2/m)
l2=(1/2)n2 √( T2/m) -----1
[m is constant because both wires are made of same material]
Divide (i) by (ii)
l1/l2=n2/n2 √ (T1/T2)
=(2/1) √(1/8)
l1/l2=0.707 : 1
Answered by
5
Given ;
n 1/n 2 = 1/ 2
T 1/T 2 = 1/ 8
n =( 1/2l)√( T/2m)
n1=(1/2l1) √ (T1/m)
l1=1/2n1 √ (T1/m) ----1
for second case
n2=(/2) l2√(T2/m)
l2=(1/2)n2 √( T2/m) -----1
[m is constant because both wires are made of same material]
Divide (i) by (ii)
l1/l2=n2/n2 √ (T1/T2)
=(2/1) √(1/8)
l1/l2=0.707 : 1
n 1/n 2 = 1/ 2
T 1/T 2 = 1/ 8
n =( 1/2l)√( T/2m)
n1=(1/2l1) √ (T1/m)
l1=1/2n1 √ (T1/m) ----1
for second case
n2=(/2) l2√(T2/m)
l2=(1/2)n2 √( T2/m) -----1
[m is constant because both wires are made of same material]
Divide (i) by (ii)
l1/l2=n2/n2 √ (T1/T2)
=(2/1) √(1/8)
l1/l2=0.707 : 1
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