Question

Simple Harmonic Motion The model$\mathit{m}{\mathit{x}}^{\mathbf{\backslash }\mathbf{c}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{\backslash }\mathbf{c}\mathbf{e}\mathbf{n}\mathbf{t}}\mathbf{+}\mathit{k}\mathit{x}\mathbf{=}\mathbf{0}$ for simple harmonic motion, discussed in Section 5.1, can be related to Example 2 of this section.

Consider a free undamped spring/mass system for which the spring constant is, say,$\mathit{k}\mathbf{}\mathbf{=}\mathbf{}\mathbf{10}\mathit{l}\mathit{b}\mathbf{/}\mathit{f}\mathit{t}$ . Determine those masses that can be attached to the spring so that when each mass is released at the equilibrium position at$\mathit{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$ with a nonzero velocity it will then pass through the equilibrium position at second. How many times will each mass pass through the equilibrium position in the time interval ?

Short answer

How many times will each mass pass through the equilibrium position in the time intervalis$0<t<1$is$n-1$.

Step-by-step solution

To Find the initial-value problem  

The solution of the initial-value problem

$x^{\backslash \mathrm{cent}\backslash \mathrm{cent}}+{\omega }^{2}x=0,x\left(0\right)=0,x^{\backslash \mathrm{cent}}\left(0\right)=v_0$where${\omega }^2=10/m,$ is given by

$x\left(t\right)=\left(v_0/\omega \right)sin\omega t$

To satisfy the additional boundary condition $x\left(1\right)=0$we require that .

$\omega =n\pi ,n=1,2,3,L$The eigenvalues and eigenfunctions of the problem are then

Find the integrate

Using ${\omega }^2=10/m$, we find that the only masses that can pass through the equilibrium position at t = 1 are $mn=10/n^2{\pi }^2$.

Note for n = 1 the heaviest mass $m_1=10/{\pi }^2$will not pass through the equilibrium position on the interval $0<t<1$(the period of $x\left(t\right)=\left(v_0/\pi \right)sin\pi t$ is T =2 seconds, so on $0⩽t⩽1$its graph passes through x = 0 only at t = 0 and t = 1 ).

Whereas for $n>1,$masses of lighter weight will pass through the equilibrium position n = 1 times prior to passing through at t = 1 For example, if n = 2 the period of $x\left(t\right)=\left(v_0/2\pi \right)sin2\pi t$ is 1 , the mass will pass through x =0 only once ( $\left.t=\frac{1}{2}\right)$( second prior to t = 1 ; n = 3 if the period of $x\left(t\right)=\left(v_0/3\pi \right)sin3\pi t$is second, the mass will pass through x = 0 twice $\left(t=\frac{1}{3}\right.$and $t=\frac{2}{3}$prior to t = 1; and so on.

Final proof

$\mathit{n}\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}$