Question

A string with a length of 4.00 m is held under a constant tension. The string has a linear mass density of μ = 0.006 kg/m. Two resonant frequencies of the string are 400 Hz and 480 Hz. There are no resonant frequencies between the two frequencies.
(a) What are the wavelengths of the two resonant modes?
(b) What is the tension in the string?

Answer 1

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