Speed of a Stretched Spring

a) There are two ways to get the answer: by the equation of motion (finding the time for the maximum point for the derivative) and by Conservation of Energy.

b) Same hint as part (a), but use half the time found there.

c) With the velocity found in part (a), substitute the new mass.

a) To do it with an alternative way as the detailed solution, Conservation of Energy gives us the answer directly:

\begin{equation*}
E_i = E_f,
\end{equation*}

which in this case is:

\begin{equation*}
\frac{1}{2} kA^2 = \frac{1}{2}mv^2.
\end{equation*}

Solving for the velocity we have:

\begin{equation*}
v=A\sqrt{\frac{k}{m}},
\end{equation*}

which, with numerical values, becomes:

\begin{equation*}
v \approx 2.23\,\text{m/s}.
\end{equation*}

Note the equation for the velocity is the same as the Detailed solution.

b) The conservation of energy this case becomes:

\begin{equation*}
\frac{1}{2} kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} k \left(\frac{A}{2}\right)^2,
\end{equation*}

and solving for the velocity we get:

\begin{equation*}
v=A\sqrt{\frac{k}{m}}\frac{\sqrt{3}}{2}.
\end{equation*}

or with numerical values:

\begin{equation*}
v \approx 1.93 \,\text{m/s}.
\end{equation*}

Again, the equation for the velocity is the same as the Detailed solution.

c) With the equation obtained for the velocity at a), by replacing \(2m\) we obtain:

\begin{equation*}
v \approx 1.58\,\text{m/s}.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

a) The problem ask us first to find the speed of the spring at its equilibrium point. The spring describes a simple harmonic motion along the X axis of amplitude \(A=0.8\,\text{m}\) and angular frequency \(\omega\) given by

\begin{equation}
\label{omega}
\omega=\sqrt{\frac{k}{m}}.
\end{equation}

Thus we can write the position \(x\), which is the same as the deformation of the spring, for any time \(t\) as follows

\begin{equation}
\label{xt}
x(t)=A\cos(\omega t),
\end{equation}

considering that at \(t=0\) the spring is maximally stretched.

The velocity is the time derivative of the position, namely

\begin{equation}
v(t)=\frac{dx(t)}{dt},
\end{equation}

which can be calculated explicitly using equation \eqref{xt} as follows

\begin{equation}
v(t)=\frac{d(A\cos(\omega t))}{dt},
\end{equation}

\begin{equation}
\label{vt}
v(t)=-A\omega \sin(\omega t).
\end{equation}

When the spring is at equilibrium its position is \(x=0\). Thus using equation \eqref{xt} we get

\begin{equation}
0=A\cos(\omega t),
\end{equation}

which is true when \(\omega t=\frac{\pi}{2}\left(2n-1\right)\), with \(n\) an integer. Solving for \(t\) we get

\begin{equation}
t=\frac{\pi}{2\omega}\left(2n-1\right).
\end{equation}

Using the expression for \(t\) given above into equation \eqref{vt} we obtain

\begin{equation}
v(t)=-A\omega\sin\left(\omega \frac{\pi}{2\omega}\left(2n-1\right)\right),
\end{equation}

which simplifies to

\begin{equation}
v(t)= -A\omega\sin\left(\frac{\pi}{2}\left(2n-1\right)\right),
\end{equation}

\begin{equation}
\label{vt2}
v(t)=-A\omega (-1)^{n+1},
\end{equation}

where we have used in the last line the trigonometric identity \(\sin\left(\frac{\pi}{2}\left(2n-1\right)\right)=(-1)^{n+1}\) which is easily verifiable numerically. From equation \eqref{vt2} we then deduce that the velocity has a magnitude of

\begin{equation}
|v(t)|=A\omega,
\end{equation}

which after using the explicit expression in equation \eqref{omega} we get

\begin{equation}
|v(t)|=A\sqrt{\frac{k}{m}}.
\end{equation}

Using the numerical values

\begin{equation}
|v(t)|=(0.8\,\text{m})\sqrt{\frac{35\,\text{N/m}}{4.5\,\text{kg}}},
\end{equation}

\begin{equation}
|v(t)|\approx 2.23\,\text{m/s}.
\end{equation}

b) Now, we need to find the speed of the spring when it is at half its maximum displacement, namely, at \(x=\frac{A}{2}\). Thus using equation \eqref{xt} we get

\begin{equation}
\frac{A}{2}=A\cos(\omega t),
\end{equation}

which is simplified to

\begin{equation}
\frac{1}{2}=\cos(\omega t).
\end{equation}

Solving for \(t\) we get

\begin{equation}
t=\frac{1}{\omega}\arccos\left(\frac{1}{2}\right),
\end{equation}

\begin{equation}
t=\frac{1}{\omega}\frac{\pi}{3}.
\end{equation}

It is important to use the angles in radians when taking the \(\arccos\) and the other inverse trigonometric functions. Using the expression for \(t\) given above into equation \eqref{vt} we obtain

\begin{equation}
v(t)=-A\omega\sin\left(\omega \frac{1}{\omega}\frac{\pi}{6}\right),
\end{equation}

which simplifies to

\begin{equation}
v(t)= -A\omega\sin\left(\frac{\pi}{6}\right),
\end{equation}

\begin{equation}
\label{vt3}
v(t)=-A\omega \frac{\sqrt{3}}{2}.
\end{equation}

From equation \eqref{vt3} we then deduce that the velocity has a magnitude of

\begin{equation}
|v(t)|=A\omega\frac{\sqrt{3}}{2},
\end{equation}

which after using the explicit expression in equation \eqref{omega} we get

\begin{equation}
|v(t)|=A\sqrt{\frac{k}{m}}\frac{\sqrt{3}}{2}.
\end{equation}

Using the numerical values

\begin{equation}
|v(t)|=(0.8\,\text{m})\sqrt{\frac{35\,\text{N/m}}{4.5\,\text{kg}}}\frac{\sqrt{3}}{2},
\end{equation}

\begin{equation}
|v(t)|\approx 1.93\,\text{m/s}.
\end{equation}

c) If the mass doubles with the same initial conditions we have that the amplitude does not change since it only depends on the initial position and velocity (neither of them changing). The velocity is given by equation \eqref{vt} with a new angular frequency \(\omega'\), namely

\begin{equation}
v(t)=-A\omega'\sin(\omega't).
\end{equation}

Because the sinusoidal function is bounded by \(-1\leq \sin(\omega't)\leq 1\) then the maximum magnitude of the velocity is

\begin{equation}
\label{vmag}
|v(t)|_{\text{max}}=A\omega',
\end{equation}

where \(\omega'\) is defined as in \eqref{omega} with the new mass \(m'\) which is the double of \(m\). Thus equation \eqref{vmag} can be written with the help of equation \eqref{omega} as

\begin{equation}
|v(t)|_{\text{max}}=A\sqrt{\frac{k}{m'}}.
\end{equation}

Using the numerical given values

\begin{equation}
|v(t)|_{\text{max}}=(0.8\,\text{m})\sqrt{\frac{35\,\text{N/m}}{9.0\,\text{kg}}},
\end{equation}

\begin{equation}
|v(t)|_{\text{max}}\approx 1.58\,\text{m/s}.
\end{equation}