Answer:
a) λn = 1.6m , λ(n+1) = 1.33
b) F(t) = 2457.6N
Explanation:
L = 4.0m
μ = 0.006kg/m
F1 =F(n) = 400Hz
F2 = F(n+1) = 480Hz
The natural frequencies of normal nodes for waves on a string is
F(n) = n(v / 2L)
F(n) = n / 2L √(F(t) / μ)
where F(t) = tension acting on the string
the speed (v) on a wave is dependent on the tension acting on the string F(t) and mass per unit length μ
v = √(F(t) / μ)
F(n) = frequency of the nth normal node
F(n+1) = frequency of the successive normal node.
frequency of thr nth normal node F(n) = n(v / 2L).......equation (i)
frequency of the (n+1)th normal node F(n+1) = (n+1) * (v / 2L).....equation (ii)
dividing equation (ii) by (i)
F(n+1) / F(n) = [(n+1) * (v / 2L)] / [n (v / 2L)]
F(n+1) / F(n) = (n + 1) / n
F(n +1) / Fn = 1 + (1/n)
1/n = [F(n+1) / Fn] - 1
1/n = (480 / 400) - 1
1/n = 1.2 - 1
1 / n = 0.2
n = 1 / 0.2
n = 5
the wavelength of the resonant nodes (5&6) nodes are
λ = 2L / n
λn = (2 * 4) / 5
λn = 8 / 5
λn = 1.6
λ(n+1) = (2 * 4) / (5 +1)
λ(n+1) = 8 / 6
λ(n+1) = 1.33m
b.
The tension F(t) acting on the string is
v = √(F(t) / μ)
v² = F(t) / μ
F(t) = μv²
but Fn = n(v / 2L)
nv = 2F(n)L
v = 2F(n)L / n
v = (2 * 400 * 4) / 5
v = 3200 / 5
v = 640m/s
substituting v = 640m/s into F(t) = v²μ
F(t) =(640)² * 0.006
F(t) = 2457.6N
