MHB Find Spring Constant \(k\) & Mass \(m\) for Natural Frequency

[Dustinsfl]

The natural frequency of a spring-mass system is found to be 2Hz. When an additional mass of 1kg is added to the original mass \(m\), the natural frequency is reduced to 1Hz. Find the spring constant \(k\) and mass \(m\).

Since the natural frequency is 2Hz, we have that \(\omega_n = 4\pi = \sqrt{\frac{k}{m}}\quad (1)\).

When 1kg is added, we have \(\omega_n = 2\pi = \sqrt{\frac{k}{m + 1}}\quad (2)\).

What I have is equation (1) with unknowns \(\sqrt{k}\) and \(\sqrt{m}\) if I write the equation as \(0 = \sqrt{k} - 4\pi\sqrt{m}\), and equation (2) with unknowns \(\sqrt{k}\) and \(\sqrt{m + 1}\) if I write the equation as \(0 = \sqrt{k} - 2\pi\sqrt{m + 1}\).

How can I reconcile these equations so I have two equations with the same two unknowns which will allow to solve for \(k\) and \(m\)?

 

[mathmari]

The natural frequency of a spring-mass system is found to be 2Hz. When an additional mass of 1kg is added to the original mass \(m\), the natural frequency is reduced to 1Hz. Find the spring constant \(k\) and mass \(m\).

Since the natural frequency is 2Hz, we have that \(\omega_n = 4\pi = \sqrt{\frac{k}{m}}\quad (1)\).

When 1kg is added, we have \(\omega_n = 2\pi = \sqrt{\frac{k}{m + 1}}\quad (2)\).

What I have is equation (1) with unknowns \(\sqrt{k}\) and \(\sqrt{m}\) if I write the equation as \(0 = \sqrt{k} - 4\pi\sqrt{m}\), and equation (2) with unknowns \(\sqrt{k}\) and \(\sqrt{m + 1}\) if I write the equation as \(0 = \sqrt{k} - 2\pi\sqrt{m + 1}\).

How can I reconcile these equations so I have two equations with the same two unknowns which will allow to solve for \(k\) and \(m\)?

To solve for $k$ and $m$ from the equations
$$0 = \sqrt{k} - 4\pi\sqrt{m} \tag 1$$
and $$0 = \sqrt{k} - 2\pi\sqrt{m + 1} \tag 2$$
we could do the following:

$$(1) \Rightarrow \sqrt{k}=4 \pi \sqrt{m}$$

Replacing this at $(2)$ we have the following:

$$0=4 \pi \sqrt{m}-2 \pi \sqrt{m+1} \Rightarrow 2 \pi \sqrt{m+1}=4 \pi\sqrt{m} \Rightarrow \sqrt{m+1}=2\sqrt{m} \Rightarrow \left (\sqrt{m+1} \right )^2=\left (2\sqrt{m} \right )^2 \Rightarrow m+1=4m \Rightarrow 3m=1 \Rightarrow m=\frac{1}{3}$$

Replacing this at the relation $\displaystyle{\sqrt{k}=4 \pi \sqrt{m}}$ we get:

$$\sqrt{k}=4 \pi \sqrt{\frac{1}{3}} \Rightarrow \left (\sqrt{k}\right )^2=\left (4 \pi \sqrt{\frac{1}{3}}\right )^2 \Rightarrow k=\frac{16}{3}\pi^2$$

 

[Dustinsfl]

To solve for $k$ and $m$ from the equations
$$0 = \sqrt{k} - 4\pi\sqrt{m} \tag 1$$
and $$0 = \sqrt{k} - 2\pi\sqrt{m + 1} \tag 2$$
we could do the following:

$$(1) \Rightarrow \sqrt{k}=4 \pi \sqrt{m}$$

Replacing this at $(2)$ we have the following:

$$0=4 \pi \sqrt{m}-2 \pi \sqrt{m+1} \Rightarrow 2 \pi \sqrt{m+1}=4 \pi\sqrt{m} \Rightarrow \sqrt{m+1}=2\sqrt{m} \Rightarrow \left (\sqrt{m+1} \right )^2=\left (2\sqrt{m} \right )^2 \Rightarrow m+1=4m \Rightarrow 3m=1 \Rightarrow m=\frac{1}{3}$$

Replacing this at the relation $\displaystyle{\sqrt{k}=4 \pi \sqrt{m}}$ we get:

$$\sqrt{k}=4 \pi \sqrt{\frac{1}{3}} \Rightarrow \left (\sqrt{k}\right )^2=\left (4 \pi \sqrt{\frac{1}{3}}\right )^2 \Rightarrow k=\frac{16}{3}\pi^2$$

thanks. That was pretty obvious. I was thinking about it all wrong.