Standing transverse wave

Question

The diagram represents a snapshot of a standing transverse wave on a flexible string taken when the displacement is at a maximum. The string is 1.25 m long with tension 4.00 N. The total mass of the string is 7.00 g. Find the period of the oscillation.

Answer 1

Let

L = 1.25 m be the length of the string,

F = 4.00 N be the tension on the string,

m = 7.00 g = 0.007 kg be the mass of the string,

d = m/L be linear density,

T be the period of the oscilation (to be calculated),

v be the speed of wave propagation (unknown),

λ be the wavelength (unknown).

To solve this problem you need the following identities

T = λ/v

v² = F/d

Combine them into

T = λ√(d/F) = λ√(m/LF) (1)

All the quantities above are given explicitly, except λ.

That can be observed from the picture showing 3.5 wavelength covering the length L.

So λ = L/3.5

Plug that into (1)

T = (L/3.5)√(m/LF) = √(mL/F) / 3.5

Plug in actual numbers

T = √(0.007×1.25/4) / 3.5 ≈ 0.013 s