At the end of a ride at a winter-themed amusement park, a sleigh with mass 250 kg (including two passengers) slides without friction along a horizontal, snow-covered surface. The sleigh hits one end of a light horizontal spring that obeys Hooke’s law and has its other end attached to a wall. The sleigh latches onto the end of the spring and subsequently moves back and forth in SHM on the end of the spring until a braking mechanism is engaged, which brings the sleigh to rest. The frequency of the SHM is 0.225 Hz , and the amplitude is 0.95 m. (a) What was the speed of the sleigh just before it hit the end of the spring? (b) What is the maximum magnitude of the sleigh’s acceleration during its SHM?

The velocity in S.H.M., at a given displacement is given by

$$\begin{gathered} {\mathbf{v}}_{\mathbf{x}}\mathbf{=}\mathbf{\pm }\sqrt{\frac{\mathbf{k}}{\mathbf{m}}}\sqrt{{\mathbf{A}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\pm }\mathbf{\omega }\sqrt{{\mathbf{A}}^{\mathbf{2}}\mathbf{-}{\mathbf{X}}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\pm }\mathbf{2}\mathbf{\pi f}\sqrt{{\mathbf{A}}^{\mathbf{2}}\mathbf{-}{\mathbf{X}}^{\mathbf{2}}} \end{gathered}$$

Here, k is the spring constant, m is the mass, A is the amplitude, and x is the displacement.

Given that

The mass, m =250 kg,

The frequency, f =0.225 Hz

The displacement, x=0

The amplitude, A = 0.95 m

Define the velocity in S.H.M as below.

$$\begin{gathered} {\mathrm{v}}_{\mathrm{x}}=\pm 2\pi \mathrm{f}\sqrt{{\mathrm{A}}^2-{\mathrm{x}}^2} \\ =\pm 2\mathrm{\pi f}\sqrt{{\mathrm{A}}^2-0} \\ =\pm 2\mathrm{\pi fA} \\ =\pm 2\mathrm{\pi }\times 0.225\times 0.95 \\ {\mathrm{v}}_{\mathrm{x}}=\pm 1.34\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the sleigh just before it hit the end of the spring is ${\mathrm{v}}_{\mathrm{x}}=\pm 1.34\mathrm{m}/\mathrm{s}$.

The formula of acceleration is

${\mathbf{a}}_{\mathbf{x}}\mathbf{=}\mathbf{-}\frac{\mathbf{k}}{\mathbf{m}}\mathbf{x}$

The maximum acceleration occurs at the most negative value of displacement, i.e., x=-A .

Therefore, the acceleration becomes,

$$\begin{gathered} a_{max}=-\frac{k}{m}\left(-A\right) \\ =\frac{k}{m}A \\ ={\omega }^{2}A \\ =4{\mathrm{\pi }}^2{\mathrm{f}}^2\mathrm{A} \end{gathered}$$

Substitute known values in the above equation, and you have

$$\begin{gathered} {\mathrm{a}}_{\max }=4\times {\left(3.14\right)}^2\times {\left(0.225\right)}^2\times 0.95 \\ =1.9\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the maximum magnitude of the sligh’s acceleration during its SHM is ${\mathrm{a}}_{\max }=1.9\mathrm{m}/{\mathrm{s}}^2$.