A wire having a linear density 0.1kg/m is kept under a tension of 490 n. It is observed that it resonates at a frequency of 400hz and the next higher frequency 450hz. Find the length of the wire
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Hey dear,
◆ Answer-
l = 0.7 m
◆ Explaination-
# Given-
T = 490 N
μ = 0.1 kg/m
f(n) = 400 Hz
f(n+1) = 450 Hz
# Solution-
Velocity of waves on string is given by-
v = √(T/μ)
v = √(490/0.1)
v = 70 m/s
Length of the wire cak be calculated by-
l = nλ1/2 = (n+1)λ2/2
n × 70/400 = (n+1) × 70/450
45n = 40n + 40
n = 8
Substitute this in ,
l = 8×70 / (400×2)
l = 0.7 m
Therefore, length of the wire is 0.7 m.
Hope it helps you...
◆ Answer-
l = 0.7 m
◆ Explaination-
# Given-
T = 490 N
μ = 0.1 kg/m
f(n) = 400 Hz
f(n+1) = 450 Hz
# Solution-
Velocity of waves on string is given by-
v = √(T/μ)
v = √(490/0.1)
v = 70 m/s
Length of the wire cak be calculated by-
l = nλ1/2 = (n+1)λ2/2
n × 70/400 = (n+1) × 70/450
45n = 40n + 40
n = 8
Substitute this in ,
l = 8×70 / (400×2)
l = 0.7 m
Therefore, length of the wire is 0.7 m.
Hope it helps you...
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