Question

Refer the attached picture. ​

Answer 1

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

Answer 2

 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