 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2019-resit/1/solution

Convert the fixed position to

x = 0.20 m

. The element's displacement is then

y (t) = 0.27 sin (2.4 - 500 t) mm

. Taking time derivatives at fixed position gives its transverse velocity and acceleration:

v_{y} (t) = - 135 cos (2.4 - 500 t) mm s^{- 1}, a_{y} (t) = - 67500 sin (2.4 - 500 t) mm s^{- 2} .

Equivalently the prefactors in SI units are

- 0.135 m s^{- 1}

and

- 67.5 m s^{- 2}

. The phase is measured in radians. At

t = 4 s

y (4) = 0.27 sin (- 1997.6) mm ≃ 0.118152 mm = 1.18152 \times 1 0^{- 4} m .

These are the transverse motions of the material element, not the propagation speed of the wave along the string.

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