Frequency and Period of SHM
Why this matters
Imagine you’re at a park in Boston, pushing your younger cousin, Leo, on a swing. You give him a small push, and he swings back and forth, back and forth. You notice that whether you give him a tiny push or a slightly bigger one, the time it takes for him to swing out and come back to you seems to be the same. It's like the swing has its own natural rhythm.
Now, what if an adult, who is much heavier than Leo, gets on the same swing? Or what if you could magically shorten the swing's chains? Would the rhythm change? This timing—the steady beat of an oscillation—is what physicists call the period and frequency. Understanding it is key to describing everything from a swaying skyscraper in the wind to the vibrations of a guitar string. In this lesson, we'll get the formulas to predict this rhythm perfectly.
Concept overview
flowchart TD
A[Start: Analyze the oscillating system] --> B{Is it a mass on a spring?};
B -- Yes --> C[Use Spring Period Formula];
C --> D[T = 2π * sqrt(m/k)];
D --> E[Period depends on Mass (m) and Spring Constant (k)];
E --> F[Period does NOT depend on Amplitude (A)];
B -- No --> G{Is it a mass on a string (pendulum)?};
G -- Yes --> H[Use Pendulum Period Formula];
H --> I[T = 2π * sqrt(L/g)];
I --> J[Period depends on Length (L) and Gravity (g)];
J --> K[Period does NOT depend on Mass (m) or Amplitude (A)];
F --> L[End];
K --> L[End];
G -- No --> M[Not simple harmonic motion covered in this topic];
M --> L;
Core explanation
When an object is in Simple Harmonic Motion, it follows a predictable, repeating pattern. The two most fundamental ways we measure this pattern are with period and frequency.
Period vs. Frequency: Two Sides of the Same Coin
Let's get our definitions straight first.
- Period (T)This is the time it takes to complete one full cycle of motion. The unit for period is seconds (s). If a pendulum takes 3 seconds to swing from its starting point, all the way to the other side, and back to the start, its period is
T = 3 s. - Frequency (f)This is the number of cycles completed in one second. The unit for frequency is Hertz (Hz), where
1 Hz = 1 cycle/second. If a spring bobs up and down 4 times every second, its frequency isf = 4 Hz.
They measure the same underlying rhythm, just from different perspectives. This leads to a beautifully simple relationship:
T = 1 / f
Think about it: if the period is long (it takes a lot of time for one cycle), the frequency must be low (you complete very few cycles per second). If the period is short (a cycle is super fast), the frequency must be high (you can fit many cycles into one second).
The Period of a Mass-Spring System
Now, let's look at our first classic SHM example: a mass attached to an ideal spring, sliding on a frictionless surface. What determines its period? It comes down to two things: the mass (m) and the spring's stiffness, represented by the spring constant (k).
The formula for the period of a mass-spring oscillator is:
T_s = 2π * √(m / k)Let's break this down so it feels intuitive.
- Mass (m)The mass is in the numerator, under the square root. This means if you increase the mass, you increase the period. This makes sense! A heavier block has more inertia, making it harder for the spring to pull and push. It will oscillate more slowly.
- Spring Constant (k)The spring constant is in the denominator. A higher
kmeans a stiffer spring. If you increase the spring constant, you decrease the period. This also makes sense. A stiffer spring exerts a stronger restoring force, snapping the mass back and forth more quickly.
The Period of a Simple Pendulum
Our second classic example is a simple pendulum: a mass (called a bob) hanging from a string of length l. For a pendulum, the restoring force is a component of gravity.
The formula for the period of a simple pendulum, for small angles (less than about 15°), is:
T_p = 2π * √(l / g)Let's analyze this one too.
- Length (l)The length of the string is in the numerator. If you increase the length, you increase the period. Think of the swing set at the park. A long swing has a slow, lazy rhythm. A short one swings back and forth much faster.
- Gravitational Acceleration (g)The value of
g(about 9.8 m/s² on Earth) is in the denominator. If you were to take your pendulum to the Moon, wheregis much smaller, the period would increase. The weaker gravity would provide a weaker restoring force, making the pendulum swing more slowly.
Here's the other big trap: What does the period of a simple pendulum not depend on? Two things!
- 1Mass (m)Look at the formula—mass isn't in it! This is very counter-intuitive. A 1 kg bob and a 10 kg bob on identical strings will have the exact same period. On the AP exam, this is a very common conceptual question.
- 2Amplitude (for small angles)Just like the spring, as long as you don't pull the pendulum back too far, the angle of release doesn't affect the period.
These two formulas are your essential tools for calculating the timing of any SHM system you'll see in this course. Master them, and you've mastered this topic.
Worked examples
Calculating the Rhythm of a Spring
Problem: A 0.5 kg block is attached to a horizontal spring with a spring constant of 50 N/m. The block is pulled 10 cm from its equilibrium position and released. What are the period and frequency of the oscillation?
Solution:
- 1Identify the system and your goalWe have a mass-spring system. We need to find the period (T) and frequency (f).
- 2Choose the correct formulaFor a mass-spring system, the period is given by:
T_s = 2π * √(m / k) - 3Check the variablesThe problem gives us
m = 0.5 kgandk = 50 N/m. What about the 10 cm displacement? Remember, for an ideal spring, the period does not depend on the amplitude. That 10 cm is extra information designed to see if you know this concept. We don't need it for this calculation. - 4Plug in the values and solve for the period
T_s = 2π * √(0.5 kg / 50 N/m)T_s = 2π * √(0.01 s²)T_s = 2π * (0.1 s)T_s ≈ 0.628 sSo, it takes about 0.63 seconds for the block to complete one full back-and-forth cycle.
- 5Calculate the frequencyFrequency is the inverse of the period.
f = 1 / Tf = 1 / 0.628 sf ≈ 1.59 HzThis means the block completes about 1.6 oscillations every second.
Conceptual Pendulum Problem
Problem: Priya has a pendulum with a period of 2.0 seconds. She wants to create a new pendulum with a period of 4.0 seconds. How should she change the length of the pendulum's string?
Solution:
- 1Identify the relationshipThis is a proportional reasoning problem. We need to relate the period of a pendulum (
T_p) to its length (l). The formula is:T_p = 2π * √(l / g) - 2Analyze the proportionThe period
T_pis directly proportional to the square root of the lengthl. We can write this asT ∝ √l. - 3Set up the ratioWe want to find the new length (
l_new) relative to the old length (l_old). We know the new period (T_new = 4.0 s) is double the old period (T_old = 2.0 s).T_new / T_old = √(l_new / l_old) - 4Solve for the change in length
4.0 s / 2.0 s = √(l_new / l_old)2 = √(l_new / l_old)To get rid of the square root, we square both sides of the equation:
2² = l_new / l_old4 = l_new / l_oldl_new = 4 * l_old
Try it yourself
Problem 1
A car's suspension can be modeled as a 1200 kg mass sitting on four springs, each with a spring constant k = 25,000 N/m. If the car hits a bump and starts oscillating up and down, what is the approximate period of this oscillation? (Assume the mass is distributed evenly).
Problem 2
An astronaut on a new planet wants to measure the local acceleration due to gravity, g. She builds a simple pendulum with a string of length 0.80 m. She measures the time for 10 full swings to be 25.0 seconds. What is the value of g on this planet?
Practice — 8 questions
In simple terms, this topic is about the timing of back-and-forth motion, like a swing or a bouncing spring. We'll learn how to calculate how long one cycle takes (period) and how many cycles happen per second (frequency).
T_s = 2π * √(m / k)- 7.2.A: Describe the frequency and period of an object exhibiting SHM.
- 7.2.A.1
- The period of SHM is related to the frequency f of the object's motion by the following equation: T = 1/f
- 7.2.A.1.i
- The period of an object-ideal-spring oscillator is given by the equation T_s = 2π√(m/k)
- 7.2.A.1.ii
- The period of a simple pendulum displaced by a small angle is given by the equation T_p = 2π√(l/g)
flowchart TD
A[Start: Analyze the oscillating system] --> B{Is it a mass on a spring?};
B -- Yes --> C[Use Spring Period Formula];
C --> D[T = 2π * sqrt(m/k)];
D --> E[Period depends on Mass (m) and Spring Constant (k)];
E --> F[Period does NOT depend on Amplitude (A)];
B -- No --> G{Is it a mass on a string (pendulum)?};
G -- Yes --> H[Use Pendulum Period Formula];
H --> I[T = 2π * sqrt(L/g)];
I --> J[Period depends on Length (L) and Gravity (g)];
J --> K[Period does NOT depend on Mass (m) or Amplitude (A)];
F --> L[End];
K --> L[End];
G -- No --> M[Not simple harmonic motion covered in this topic];
M --> L;
Read what Saavi narrates
Hey everyone, it's Saavi.
Imagine you’re at a park in Boston, pushing your younger cousin on a swing. You notice that whether you give him a tiny push or a slightly bigger one, the time it takes for him to swing out and come back seems to be the same. It's like the swing has its own natural rhythm. This timing—the steady beat of an oscillation—is what we call the period and frequency.
Today, we're focusing on two key measurements for any object in this kind of motion. The first is its period, which is the time for one complete cycle. The second is its frequency, which is how many cycles it completes per second.
Let's walk through an example. Say we have a 0.5 kilogram block attached to a spring with a spring constant of 50 Newtons per meter. The problem also says we pull it back 10 centimeters. What are the period and frequency?
First, we need the right formula. For a spring, the period, T, is two pi times the square root of the mass, m, divided by the spring constant, k.
Now, what about that 10 centimeter displacement? This is a classic test question. The period of a spring does not depend on how far you pull it back. So, we can ignore that number.
Let's plug in our values. The period T equals two pi times the square root of 0.5 kilograms divided by 50. That simplifies to two pi times the square root of 0.01, which is just 0.1. So, the period is two pi times 0.1, which is about 0.63 seconds. That’s the time for one full back-and-forth motion.
To find the frequency, we just take the inverse of the period. Frequency, f, is one divided by T. So, one divided by 0.63 gives us about 1.6 Hertz.
Now, here's a common mistake I see every year: students think the mass of a pendulum changes its period. It feels intuitive, right? A heavier kid on a swing should go slower. But the physics says otherwise. The formula for a pendulum's period only depends on its length and gravity, not its mass. A heavier bob experiences more gravity, but it also has more inertia, and those two effects cancel out perfectly. Don't fall for that trap on the exam.
Remember these two systems, springs and pendulums, have their own unique rules. Keep their formulas straight, and you'll do great. You've got this.
The physics governing each system is different (spring force vs. gravity), leading to different variables in their period equations. Using the wrong formula will always give the wrong answer.
Before starting a problem, write down "Spring: `T = 2π√(m/k)`" and "Pendulum: `T = 2π√(l/g)`". Explicitly circle the one you need for the problem at hand.
For both ideal springs and simple pendulums (at small angles), the period is independent of amplitude. A larger amplitude means a longer distance to travel, but also a higher average speed, and these effects cancel out.
Look at the two period formulas. Is amplitude (`A` or `x_max`) in either of them? No. Remind yourself that amplitude is often included in problems as a distractor.
The formula `T_p = 2π√(l/g)` does not include mass (`m`). While a heavier object has more inertia, it also experiences a proportionally stronger gravitational force. These two effects cancel perfectly.
Memorize this counter-intuitive fact. On a conceptual question comparing a light and heavy pendulum of the same length, their periods are identical.
The relationship between period and the variables (`m`, `k`, `l`) is not linear. If you double the length of a pendulum, the period increases by a factor of `√2`, not 2.
When you see a question asking "What happens to the period if you quadruple the mass?", always write the formula and focus on the square root. `T_new ∝ √ (4 * m_old)`, so `T_new ∝ 2 * √m_old`. The period doubles.
They are inverses and have different units. Period is time/cycle (seconds). Frequency is cycles/time (Hertz). Answering with a frequency when the question asks for a period will cost you points.
After calculating your answer, check the units. If you found `T = 2 s`, that means "2 seconds per cycle." If you found `f = 0.5 Hz`, that means "half a cycle per second." Make sure it matches what the question asked for.