Refer the attached picture.
Answers
Mass of the wire, m = 3.5 × 10–2 kg
Linear mass density, μ = m/l = 4.0 × 10-2 kg m-1
Frequency of vibration, v = 45 Hz∴ length of the wire, l = m/μ = 3.5 × 10–2 / 4.0 × 10-2 = 0.875 m
The wavelength of the stationary wave (λ) is related to the length of the wire by the relation:
λ = 2l/m
where,
n = Number of nodes in the wire
For fundamental node, n = 1:
λ = 2l
λ = 2 × 0.875 = 1.75 m
The speed of the transverse wave in the string is given as:
v = νλ= 45 × 1.75 = 78.75 m/s
(b) The tension produced in the string is given by the relation:
T = v2µ
= (78.75)2 × 4.0 × 10–2 = 248.06 N
m = 3.5 × 10–2 kg
μ = m/l = 4.0 × 10-2 kg m-1
v = 45 Hz∴ length of the wire, l = m/μ = 3.5 × 10–2 / 4.0 × 10-2 = 0.875 m
λ = 2l/m
n = Number of nodes in the wire
For fundamental node, n = 1:
λ = 2l
λ = 2 × 0.875 = 1.75 m
The speed of the transverse wave in the string is given as:
v = νλ= 45 × 1.75 = 78.75 m/s
(b) The tension produced in the string is given by the relation:
T = v2µ
= (78.75)2 × 4.0 × 10–2 = 248.06 N
answer