Question

In Exercise 83 from Section 1.6 we saw that the oscillating position of a mass hanging from the end of a spring, neglecting air resistance, is given by the following equation, where A, B, k, and mare constants:

$s\left(t\right)=A\sin \left(\sqrt{\frac{k}{m}}t\right)+B\cos \left(\sqrt{\frac{k}{m}}t\right)$

(a) Show that the function s(t) has the property that $s\text{'}\text{'}\left(t\right)+\frac{k}{m}s\left(t\right)=0$. This is the differential equation for the spring motion, an equation involving derivatives that describes the motion of the bob on the end of the spring.

(b) Suppose the spring is released from an initial position of $s_0$and with an initial velocity of $v_0$. Show that $A=v_0\sqrt{\frac{m}{k}};B=s_0$

Short answer

(a) The function $s\left(t\right)=A\sin \left(\sqrt{\frac{k}{m}}t\right)+B\cos \left(\sqrt{\frac{k}{m}}t\right)$has the property that, $s\text{'}\text{'}\left(t\right)+\frac{k}{m}s\left(t\right)=0$

(b) If the spring is released from an initial position of $s_0$and with an initial velocity of $v_0$, then $A=v_0\sqrt{\frac{m}{k}};B=s_0$

Step-by-step solution

Part (a) Step 1. Given Information.

The function:

$s\left(t\right)=A\sin \left(\sqrt{\frac{k}{m}}t\right)+B\cos \left(\sqrt{\frac{k}{m}}t\right)$

Part (a) Step 2. Find the first and second derivative of the given function.

Find the first derivative:

$$\begin{gathered} s\left(t\right)=A\sin \left(\sqrt{\frac{k}{m}}t\right)+B\cos \left(\sqrt{\frac{k}{m}}t\right) \\ s\text{'}\left(t\right)=A\cos \left(\sqrt{\frac{k}{m}}t\right)\left(\sqrt{\frac{k}{m}}\right)+B(-\sin )\left(\sqrt{\frac{k}{m}}t\right)\left(\sqrt{\frac{k}{m}}\right) \\ =A\left(\sqrt{\frac{k}{m}}\right)\cos \left(\sqrt{\frac{k}{m}}t\right)-B\left(\sqrt{\frac{k}{m}}\right)\sin \left(\sqrt{\frac{k}{m}}t\right) \end{gathered}$$

Find the second derivative from the first derivative.

$$\begin{gathered} s\text{'}\text{'}\left(t\right)=A\left(\sqrt{\frac{k}{m}}\right)(-\sin )\left(\sqrt{\frac{k}{m}}t\right)-B\left(\sqrt{\frac{k}{m}}\right)\cos \left(\sqrt{\frac{k}{m}}t\right) \\ =-A\left(\sqrt{\frac{k}{m}}\right)\sin \left(\sqrt{\frac{k}{m}}t\right)-B\left(\sqrt{\frac{k}{m}}\right)\cos \left(\sqrt{\frac{k}{m}}t\right) \end{gathered}$$

Part (a) Step 3. Multiply s(t) by km.

Multiplying the given function with $\frac{k}{m}$,

$$\begin{gathered} s\left(t\right)\left(\frac{k}{m}\right)=[A\sin (\sqrt{\frac{k}{m}}t)+B\cos (\sqrt{\frac{k}{m}}t\left)\right]\left(\frac{k}{m}\right) \\ =\left[A\right(\frac{k}{m})\sin (\sqrt{\frac{k}{m}}t)+B(\frac{k}{m})\cos (\sqrt{\frac{k}{m}}t\left)\right] \end{gathered}$$

Part (a) Step 4. Add the obtained equations.

$$\begin{gathered} s\text{'}\text{'}\left(t\right)+\left(\frac{k}{m}\right)s\left(t\right)=[-A(\sqrt{\frac{k}{m}}\left)\sin \right(\sqrt{\frac{k}{m}}t)-B(\sqrt{\frac{k}{m}}\left)\cos \right(\sqrt{\frac{k}{m}}t\left)\right]+\left[A\right(\frac{k}{m})\sin (\sqrt{\frac{k}{m}}t)+B(\frac{k}{m})\cos (\sqrt{\frac{k}{m}}t\left)\right] \\ =\left(\sqrt{\frac{k}{m}}\right)\sin \left(\sqrt{\frac{k}{m}}\right)(A-A)+B\left(\sqrt{\frac{k}{m}}\right)\cos \left(\sqrt{\frac{k}{m}}t\right)(B-B) \\ =0 \end{gathered}$$

Thus the given function has the property that

Part (b) Step 1. Find the point A.

It is given that the spring is released from an initial position of $s_0$with the initial velocity $v_0$.

$$\begin{gathered} s\text{'}\left(0\right)=v_0 \\ So, \\ s\text{'}\left(0\right)=A\left(\sqrt{\frac{k}{m}}\right)\cos \left(\sqrt{\frac{k}{m}}\right(0\left)\right)-B\left(\sqrt{\frac{k}{m}}\right)\sin \left(\sqrt{\frac{k}{m}}\right(0\left)\right) \\ {v}_0=A\left(\sqrt{\frac{k}{m}}\right)\cos \left(0\right)-B\left(\sqrt{\frac{k}{m}}\right)\sin \left(0\right) \\ =A\left(\sqrt{\frac{k}{m}}\right) \\ A=v_0\left(\sqrt{\frac{m}{k}}\right) \end{gathered}$$

Part (b) Step 2. Find B.

It is given that the spring is released from an initial position of $s_0$with the initial velocity $v_0$.

We know that,

$$\begin{gathered} s\left(0\right)=s_0 \\ s\left(0\right)=A\sin \left(\sqrt{\frac{k}{m}}\right(0\left)\right)+B\cos \left(\sqrt{\frac{k}{m}}\right(0\left)\right) \\ {s}_0=A\sin \left(0\right)+B\cos \left(0\right) \\ {s}_0=B \end{gathered}$$

Hence it has been proved.