Question

A 20-kilogram mass is attached to a spring. If the frequency of simple harmonic motion is $\frac{\mathbf{2}}{\mathbf{\pi }}$cycles/s, what is the spring constant? What is the frequency of simple harmonic motion if the original mass is replaced with an 80-kilogram mass?

Short answer

So, the required frequency is$\mathit{f}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}$ .

Step-by-step solution

Definition of Newton’s second law

Newton's second law tells us exactly how much an object will accelerate for a given net force. To be clear, is the acceleration of the object, is the net force on the object, and is the mass of the object.

Solve for the spring constant

Suppose that a 20 -kilogram mass is attached to a spring and the frequency of simple harmonic motion is$\mathbf{2}\mathbf{/}\mathit{\pi }$cycles/sec.

It is needed to determine the spring constant $\mathit{k}$.

After a mass $\mathit{m}\mathbf{=}\mathbf{20}\mathbf{}\mathbf{\text{kg}}$is attached to a spring, it stretches the spring by an amount and attains a position of equilibrium at which its weight $\mathit{W}\mathbf{=}\mathit{m}\mathit{g}$is balanced by the restoring force$\mathbf{\text{ks}}$.

If the mass is displaced by an amount $\mathit{x}$from its equilibrium position, the restoring force of the spring is then $\mathit{k}\mathbf{(}\mathit{x}\mathbf{+}\mathit{s}\mathbf{)}$.

Use of Newton’s second law

Equate Newton's second law with the net, or resultant, force of the restoring force and the weight:

$\mathbf{20}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{=}\mathbf{-}\mathit{k}\mathbf{(}\mathit{x}\mathbf{+}\mathit{s}\mathbf{)}\mathbf{+}\mathit{m}\mathit{g}$

$\mathbf{=}\mathbf{-}\mathit{k}\mathit{x}\mathbf{+}\mathit{m}\mathit{g}\mathbf{-}\mathit{k}\mathit{s}$

At the equilibrium position$\mathit{m}\mathit{g}\mathbf{=}\mathit{k}\mathit{s}$.

$\mathbf{20}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{=}\mathbf{-}\mathit{k}\mathit{x}\mathbf{}\mathbf{}\mathbf{}\mathbf{\dots }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

The negative sign in (1) indicates that the restoring force of the spring acts opposite to the direction of motion.

Find second order equation

By dividing (1) by 20 , we obtain the second order differential equation

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{k}}{\mathbf{20}}\mathit{x}\mathbf{=}\mathbf{0}$

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}{\left(\sqrt{\frac{k}{20}}\right)}^{\mathbf{2}}\mathit{x}\mathbf{=}\mathbf{0}\mathbf{\dots }\mathbf{\dots }\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

To solve the differential equation$\mathbf{\left(}\mathbf{2}\mathbf{\right)}$ , consider its auxiliary equation.

${\mathit{m}}^{\mathbf{2}}\mathbf{+}{\left(\sqrt{\frac{k}{20}}\right)}^{\mathbf{2}}\mathbf{=}\mathbf{0}$

The solutions of the auxiliary equation ${\mathit{m}}^{\mathbf{2}}\mathbf{+}{\left(\sqrt{\frac{k}{20}}\right)}^{\mathbf{2}}\mathbf{=}\mathbf{0}$are complex numbers${\mathit{m}}_{\mathbf{1}}\mathbf{=}\sqrt{\frac{\mathbf{k}}{\mathbf{20}}}\mathit{i}$and${\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\sqrt{\frac{\mathbf{k}}{\mathbf{20}}}\mathit{i}$

So, the general solution of $\mathbf{\left(}\mathbf{1}\mathbf{\right)}$is $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{cos}\left(\sqrt{\frac{k}{20}}\right)\mathbf{t}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{sin}\left(\sqrt{\frac{k}{20}}\right)\mathbf{t}$.

Frequency of the Simple Harmonic Motion

Recall the result that, frequency of the simple harmonic motion described by the equation $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}{\mathit{\omega }}^{\mathbf{2}}\mathit{x}\mathbf{=}\mathbf{0}$whose general solution is$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathit{\omega }\mathit{t}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{\omega }\mathit{t}$is $\mathit{f}\mathbf{=}\frac{\mathbf{\omega }}{\mathbf{2}\mathbf{\pi }}$number of cycles complete each second.

Compare the solution $\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\left(\sqrt{\frac{k}{20}}\right)\mathit{t}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\left(\sqrt{\frac{k}{20}}\right)\mathit{t}$with $\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathit{\omega }\mathit{t}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{\omega }\mathit{t}$, we get $\mathit{\omega }\mathbf{=}\sqrt{\frac{\mathbf{k}}{\mathbf{20}}}$

So, frequency of the simple harmonic motion described by the equation (2) is $\mathit{f}\mathbf{=}\frac{\left(\sqrt{\frac{k}{20}}\right)}{\mathbf{2}\mathbf{\pi }}$Replace $\mathit{\omega }$with $\sqrt{\frac{\mathbf{k}}{\mathbf{20}}}$in $\mathit{f}\mathbf{=}\frac{\mathbf{\omega }}{\mathbf{2}\mathbf{\pi }}$.

Solve for spring constant

From the data, we have frequency of simple harmonic motion is$\frac{\mathbf{2}}{\mathbf{\pi }}$cycles/sec .

So, we get

$\frac{\left(\sqrt{\frac{k}{20}}\right)}{\mathbf{2}\mathbf{\pi }}\mathbf{=}\frac{\mathbf{2}}{\mathbf{\pi }}$

$\mathbf{\{}\frac{\mathbf{k}}{\frac{\mathbf{k}}{\mathbf{0}}}\mathbf{=}\mathbf{4}$

$\frac{\mathbf{k}}{\mathbf{20}}\mathbf{=}\mathbf{16}$

$\mathit{k}\mathbf{=}\mathbf{320}\mathbf{}\mathbf{\text{N}}\mathbf{/}\mathbf{\text{m}}$

Hence, the spring constant $\mathit{k}\mathbf{=}\mathbf{320}\mathbf{}\mathbf{\text{N}}\mathbf{/}\mathbf{\text{m}}$.

Solve for general solution

Suppose that, original mass 20 -kilograms replaced with 80 -kilograms.

Then the simple harmonic motion described by the equation $\mathbf{80}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{-}\mathbf{320}\mathbf{x}$.

Here spring constant$\mathit{k}\mathbf{=}\mathbf{320}\mathbf{}\mathbf{\text{N}}\mathbf{/}\mathbf{\text{m}}$

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}\mathbf{4}\mathit{x}\mathbf{=}\mathbf{0}$

To solve the differential equation (3), consider its auxiliary equation.

${\mathit{m}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{=}\mathbf{0}$

The solutions of the auxiliary equation ${\mathit{m}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{=}\mathbf{0}$are real numbers${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathit{i}$and ${\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{2}\mathit{i}$.

So, the general solution of $\mathbf{\left(}\mathbf{3}\mathbf{\right)}$is $\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{t}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathbf{2}\mathit{t}$.

Solve for frequency of Simple Harmonic Motion

Recall the result that, frequency of the simple harmonic motion described by the equation $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}{\mathit{\omega }}^{\mathbf{2}}\mathit{x}\mathbf{=}\mathbf{0}$.

Whose general solution is$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathit{\omega }\mathit{t}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{\omega }\mathit{t}$is$\mathit{f}\mathbf{=}\mathit{\omega }\mathbf{/}\mathbf{2}\mathit{\pi }$number of cycles complete each second.

Compare the solution $\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{t}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathbf{2}\mathit{t}$with $\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathit{\omega }\mathit{t}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{\omega }\mathit{t}$, we get $\mathit{\omega }\mathbf{=}\mathbf{2}$.

So, frequency of the simple harmonic motion described by the equation (3) is $\mathit{f}\mathbf{=}\frac{\mathbf{2}}{\mathbf{2}\mathbf{\pi }}$Replace$\mathit{\omega }$with 2 in $\mathit{f}\mathbf{=}\frac{\mathbf{\omega }}{\mathbf{2}\mathbf{\pi }}$$\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}$cycles/second

Hence, frequency of the simple harmonic motion when 80 -kilograms mass is attached to a spring is cycles/second$\mathit{f}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}$.