Frequency & Period of SHM (College Board AP® Physics 1: Algebra-Based): Revision Note

The horizontal object-ideal spring system above has two springs attached in series to an object with mass $m$. The springs have spring constants $k_x$ and $k_y$.

An effective spring constant is the spring constant of a single spring that would behave the same as the two springs combined.

You may assume that each spring experiences the same force.

Derive an expression for the period of the system in terms of $m$, $k_x$ and $k_y$ by finding the system's effective spring constant.

Answer:

Step 1: Analyze the scenario

This system is very similar to a standard horizontal object-ideal spring system, however there are now two springs instead of one.

To answer the question, find an expression for the effective spring constant. The question states that force is evenly distributed between the springs. This means the same value for force can be used for both springs.

Step 2: Find the effective spring constant of this system

• When the object is at a displacement $\mathrm{\Delta }x$ from the equilibrium position, this displacement is the sum of the extension of each spring:

$\mathrm{\Delta }x=\mathrm{\Delta }{x}_x+\mathrm{\Delta }{x}_y$

• Where:

  • $\mathrm{\Delta }{x}_x$ is the extension of spring x, measured in $\mathrm{m}$

  • $\mathrm{\Delta }{x}_y$ is the extension of spring y, measured in $\mathrm{m}$

• The extensions can be written in terms of force and spring constants using Hooke's law:

$F_x=-k_x\mathrm{\Delta }{x}_x$

$\mathrm{\Delta }{x}_x=-\frac{F_x}{k_x}$

$\mathrm{\Delta }{x}_y=-\frac{F_y}{k_y}$

• Recall that force is evenly distributed, so the forces exerted by each spring are equal:

$F_x=F_y=F$

$\mathrm{\Delta }x=\mathrm{\Delta }{x}_x+\mathrm{\Delta }{x}_y=\left(-\frac{F}{k_x}\right)+\left(-\frac{F}{k_y}\right)$

• Recall Hooke's law is:

$F=-k\mathrm{\Delta }x$

• Write this equation in the format of Hooke's law for a single spring:

$\mathrm{\Delta }x=-\left(\frac{1}{k_x}+\frac{1}{k_y}\right)F$

$F=-\left(\frac{1}{\frac{1}{k_x}+\frac{1}{k_y}}\right)\mathrm{\Delta }x=-\left(\frac{k_x{k}_y}{k_x+k_y}\right)\mathrm{\Delta }x$

• From this equation, it can be determined that the effective spring constant, $k_e$, is:

$k_e=\frac{k_x{k}_y}{k_x+k_y}$

Step 3: Substitute this into the equation for the period of an object-ideal spring system

• Find the equation for the period of this system from the equation sheet:

$T_s=2\mathrm{\pi }\sqrt{\frac{m}{k}}$

• Substitute $k_e$ in place of $k$ in this equation to produce the final equation:

$T_s=2\mathrm{\pi }\sqrt{\frac{m}{\frac{k_x{k}_y}{k_x+k_y}}}$

• This can be simplified to the following:

$T_s=2\mathrm{\pi }\sqrt{\frac{m\left(k_x+k_y\right)}{k_x{k}_y}}$

This equation for this system's period appears on your equation sheet:

$T_s=2\mathrm{\pi }\sqrt{\frac{m}{k}}$

You do not need to know the derivation. You may, however, need to rearrange it to find frequency or use it to derive new equations.

Additionally, this equation looks similar to the period for a pendulum, make sure you know which equation applies to which system. This equation features the spring constant, which may help to remind you.

Which equation correctly shows the frequency of a pendulum of length $l$ oscillating through a small angle?

A: $f_p=2\mathrm{\pi }\sqrt{\frac{g}{l}}$

B: $f_p=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{l}{g}}$

C: $f_p=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{g}{l}}$

D: None of the above

Answer: C

Step 1: Analyze the scenario

• A pendulum of length $l$ is oscillating through a small angle

• This means it can be modeled as SHM

Step 2: Recall the relevant equations

• From the equation sheet, the relevant equation for a pendulum's period is:

$T_p=2\mathrm{\pi }\sqrt{\frac{l}{g}}$

• Recall that frequency is the reciprocal of period:

$f=\frac{1}{T}$

Step 3: Apply the specific conditions

• Substitute the period of a pendulum into this equation for frequency:

$f_p=\frac{1}{2\mathrm{\pi }\sqrt{\frac{l}{g}}}$

• Simplify this expression:

$f_p=\frac{1}{2\mathrm{\pi }}·\frac{1}{\sqrt{{\displaystyle \frac{l}{g}}}}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{g}{l}}$

• Therefore the correct answer is C

Again, this equation is given on the equation sheet and you will not be expected to derive it in your exam. You may, however, be required to demonstrate that the restoring torque in this system arises from a component of weight acting perpendicular to the string.