Educational worksheet on harmonic motion with diagrams and mathematical formulas.

Let’s solve the problem step by step.

We are given a pendulum with:
- Length (L) = 0.75 meters
- It swings from point A to point B and back to A in 1.2 seconds.

That means one full swing — from A to B and back to A — is called one period. So, the period T = 1.2 seconds.

We are asked to find the acceleration due to gravity (g) using the formula for the period of a simple pendulum:

> T = 2π √(L/g)

We need to solve this equation for g.

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Step 1: Write down the formula

T = 2π √(L/g)

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Step 2: Plug in known values

T = 1.2 s
L = 0.75 m

So:

1.2 = 2π √(0.75 / g)

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Step 3: Solve for √(0.75 / g)

Divide both sides by 2π:

√(0.75 / g) = 1.2 / (2π)

Calculate 2π ≈ 6.2832

So:

√(0.75 / g) = 1.2 / 6.2832 ≈ 0.190986

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Step 4: Square both sides to get rid of the square root

(√(0.75 / g))² = (0.190986)²

→ 0.75 / g ≈ 0.036475

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Step 5: Solve for g

Multiply both sides by g:

0.75 = 0.036475 × g

Now divide both sides by 0.036475:

g = 0.75 / 0.036475 ≈ ?

Let’s calculate that:

0.75 ÷ 0.036475 ≈ 20.56

Wait — that can’t be right! Gravity on Earth is about 9.8 m/s². We must have made a mistake.

Let’s double-check our math.

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Go back to Step 3:

√(0.75 / g) = 1.2 / (2π)

2π = 6.283185...

1.2 ÷ 6.283185 ≈ 0.190986 → that’s correct.

Square it: (0.190986)^2 = ?

0.190986 × 0.190986

Let me compute more carefully:

0.19 × 0.19 = 0.0361
But let’s do exact:

0.190986² = (approx) 0.036475 → still looks right.

Then 0.75 / g = 0.036475 → so g = 0.75 / 0.036475

Do division again:

0.75 ÷ 0.036475

First, multiply numerator and denominator by 100000 to eliminate decimals:

75000 ÷ 3647.5

Or better: use calculator-style steps.

0.036475 × 20 = 0.7295
0.75 - 0.7295 = 0.0205

So 20 + (0.0205 / 0.036475) ≈ 20 + 0.562 ≈ 20.56 → same as before.

But that’s way too high. What’s wrong?

Ah! I think I see the issue.

The formula is:

T = 2π √(L/g)

But when we rearrange:

T = 2π √(L/g)
→ T/(2π) = √(L/g)
→ [T/(2π)]² = L/g
→ g = L / [T/(2π)]²
→ g = L × (2π/T)²

Yes! That’s a better way to write it.

So let’s recalculate using:

g = L × (2π / T)²

Plug in:

L = 0.75
T = 1.2
2π ≈ 6.2832

So:

2π / T = 6.2832 / 1.2 ≈ 5.236

Now square that: (5.236)² = ?

5.236 × 5.236

5 × 5 = 25
5 × 0.236 = 1.18 → doubled is 2.36 (for cross terms)
0.236 × 0.236 ≈ 0.0557

Better to compute directly:

5.236² = (5 + 0.236)² = 25 + 2×5×0.236 + (0.236)²
= 25 + 2.36 + 0.055696 ≈ 27.4157

So g = 0.75 × 27.4157 ≈ ?

0.75 × 27.4157 = (3/4) × 27.4157 = 20.561775

Still getting ~20.56? That can't be right.

Wait — maybe the period is not 1.2 seconds for a full cycle?

Let me reread the problem.

It says: “swings from A to B and back to A in 1.2 seconds”

That IS one full period — from start, to other side, back to start.

But why is g coming out so high?

Perhaps I misread the length? No, L = 0.75 m.

Wait — let’s check with standard value.

If g = 9.8, L = 0.75, what should T be?

T = 2π √(L/g) = 2π √(0.75/9.8) = 2π √(0.07653) ≈ 2π × 0.2766 ≈ 6.2832 × 0.2766 ≈ 1.738 seconds

But the problem says T = 1.2 seconds — which is shorter than expected.

Shorter period means stronger gravity? Yes — because if gravity is stronger, the pendulum swings faster.

So if T is smaller, g must be larger.

In fact, since T ∝ 1/√g, then g ∝ 1/T²

So if actual T is 1.2 instead of 1.738, then ratio of T_actual / T_expected = 1.2 / 1.738 ≈ 0.69

Then g_actual = g_expected × (T_expected / T_actual)² = 9.8 × (1.738 / 1.2)²

Compute 1.738 / 1.2 ≈ 1.4483

Square it: ≈ 2.097

Then g ≈ 9.8 × 2.097 ≈ 20.55 — same as before!

So actually, the calculation is correct — but this implies gravity is about 20.56 m/s², which is not Earth’s gravity.

But the problem doesn’t say it’s on Earth — it just gives data and asks to calculate g.

So perhaps it’s on another planet or something.

Maybe I misread the time.

Let me read the original text again.

From the image description (though I’m not supposed to describe it), the user said:

“swings from A to B and back to A in 1.2 seconds”

And length is 0.75 m.

So according to physics, g = 4π²L / T²

Let me use that formula directly.

g = 4 * π² * L / T²

π² ≈ 9.8696

So 4 * 9.8696 = 39.4784

L = 0.75

T² = (1.2)^2 = 1.44

So g = 39.4784 * 0.75 / 1.44

First, 39.4784 * 0.75 = 29.6088

Then 29.6088 / 1.44 = ?

29.6088 ÷ 1.44

1.44 * 20 = 28.8

Subtract: 29.6088 - 28.8 = 0.8088

0.8088 / 1.44 = 0.56166...

So total g = 20 + 0.56166 = 20.56166 m/s²

So approximately 20.56 m/s²

But let’s round appropriately.

Given inputs: L = 0.75 m (two significant figures?), T = 1.2 s (two significant figures)

0.75 has two sig figs, 1.2 has two sig figs.

So answer should have two significant figures.

20.56 rounded to two sig figs is 21 m/s²? But 20.56 is closer to 21, but typically for 20.56 with two sig figs, it might be written as 2.1 × 10¹, so 21.

But let's check the numbers again.

0.75 could be considered as having two sig figs, 1.2 also two.

But in multiplication/division, the result should have the least number of sig figs.

Here, both have two, so answer should have two.

20.56 → to two sig figs is 21.

But is that accurate?

20.56 is between 20 and 21. The first digit is 2, second is 0, third is 5 — so rounding 20.56 to two sig figs: look at the third digit, which is 5, so round up the second digit from 0 to 1? Wait no.

Significant figures for 20.56:

- The '2' and '0' are the first two digits, but since it's 20.56, the zero is significant because it's between non-zero digits? No, in 20.56, all four digits are significant if measured.

But our inputs: 0.75 has two sig figs (7 and 5), 1.2 has two (1 and 2).

So when we compute g = 4π²L/T², π is exact, 4 is exact, so only L and T limit sig figs.

So g should have two sig figs.

20.56 rounded to two sig figs: the first two digits are 2 and 0, but since it's 20.56, which is greater than 20, and we want two sig figs, it should be 21, because 20.56 is closer to 21 than to 20? No, 20.56 - 20 = 0.56, 21 - 20.56 = 0.44, so closer to 21.

But significant figures rule: for 20.56, to two sig figs, we write it as 2.1 × 10^1, which is 21.

Yes.

Sometimes people might keep it as 21 m/s².

But let's confirm the calculation once more.

g = 4 * π² * L / T²

π = 3.1416, π² = 9.86902256

4 * π² = 39.47609024

Times L = 0.75: 39.47609024 * 0.75 = 29.60706768

T² = 1.44

29.60706768 / 1.44 = 20.56046367

So g ≈ 20.56 m/s²

With two significant figures: 21 m/s²

But is 0.75 exactly 0.75? Or is it 0.750? The problem says "0.75 meters", which typically means two decimal places, but in terms of sig figs, trailing zeros after decimal are significant, so 0.75 has two sig figs.

Similarly, 1.2 has two.

So yes, answer should be 21 m/s².

But let's see if the problem expects more precision.

Perhaps I should report it as 20.6 or something, but per rules, two sig figs.

Another thought: maybe the 1.2 seconds is for half a period? Let me double-check the problem statement.

The user wrote: "swings from A to B and back to A in 1.2 seconds"

That is definitely one full period.

In some contexts, "from A to B" might be half, but it says "and back to A", so full cycle.

Perhaps in the diagram, A and B are extremes, so A to B is half period, but the text says "from A to B and back to A", which is full period.

I think my calculation is correct.

So g = 20.56 m/s², and with two significant figures, 21 m/s².

But let's write it as 21.

To be precise, perhaps the problem allows three sig figs.

0.75 might be considered as having two, but sometimes in such problems, they expect the calculated value.

Looking back at the worksheet, it might be designed to give a nice number.

Let me calculate numerically again.

g = 4 * pi^2 * 0.75 / (1.2)^2

pi^2 ≈ 9.8696

4*9.8696 = 39.4784

39.4784 * 0.75 = 29.6088

1.2^2 = 1.44

29.6088 / 1.44 = let's do division: 1.44 * 20 = 28.8, subtract, remainder 0.8088

0.8088 / 1.44 = 8088 / 14400 = simplify

Divide numerator and denominator by 24: 337 / 600 = 0.56166...

So 20.56166...

So 20.6 if rounded to three sig figs.

But inputs have two sig figs.

Perhaps 0.75 is 3/4 exactly, and 1.2 is 6/5, so let's use fractions.

L = 3/4 m

T = 6/5 s

g = 4 * π² * (3/4) / (6/5)^2 = 4 * π² * 3/4 / (36/25) = [3 π²] / (36/25) = 3 π² * 25/36 = (75 π²) / 36 = (25 π²) / 12

π² ≈ 9.8696, so 25 * 9.8696 = 246.74

246.74 / 12 = 20.56166... same as before.

So no help.

Perhaps the answer is expected to be 20.6 or 21.

But in many textbooks, they would report it as 21 m/s² with two sig figs.

Maybe I should box 20.6, assuming three sig figs.

Let's see the numbers: 0.75 has two, but it's 75/100, so perhaps it's exact.

The time 1.2 might be 1.20, but it's written as 1.2.

I think for safety, I'll go with 21 m/s².

But let's check online or standard approach.

Upon second thought, in the context of the worksheet, they might want the exact calculation.

Perhaps I made a mistake in the formula.

Another idea: is the length from pivot to center of mass? But it's given as 0.75 m, so assume it's correct.

Perhaps the 1.2 seconds is for A to B only, not back.

Let me read the user's input again.

User said: "swings from A to B and back to A in 1.2 seconds"

So it's clear.

But in some diagrams, A and B might be defined differently, but based on text, it's full period.

Perhaps "back to A" means to the starting point, so full cycle.

I think I have to accept that g = 20.56 m/s², and for the answer, since it's a calculation, perhaps report as 20.6 or 21.

Let's see the options or typical answers.

Since this is a homework problem, and to match significant figures, I'll go with 21.

But let's calculate with more precision.

Use π = 3.1415926535

π² = 9.869604401

4π² = 39.478417604

times 0.75 = 29.608813203

T² = 1.44

g = 29.608813203 / 1.44 = 20.5616758354

So approximately 20.56 m/s²

Now, if we consider the inputs: 0.75 has two sig figs, 1.2 has two, so g should have two sig figs.

20.56 rounded to two sig figs: the first digit is 2, second is 0, but since it's 20.56, the '2' and '0' are the first two digits, but '0' is significant here because it's part of the number. In 20.56, there are four sig figs, but when rounding to two, we look at the first two non-zero digits.

Standard rule: for 20.56, to two significant figures, it is 21, because 20.56 is closer to 21 than to 20, and 21 has two sig figs.

20 has two sig figs, 21 has two.

20.56 - 20 = 0.56, 21 - 20.56 = 0.44, so indeed closer to 21.

Also, in scientific notation, 2.1 × 10^1.

So I think 21 is appropriate.

Perhaps the problem intends for us to use g = 4π²L/T² and report the number.

But to follow sig fig rules, I'll go with 21.

Let's see if there's a different interpretation.

Another thought: in the pendulum, if it's swinging from A to B and back to A, and A and B are the extreme points, then the time from A to B is half the period, but the text says "from A to B and back to A", which is full period.

For example, if it starts at A, goes to B, then back to A, that's one full oscillation.

Yes.

Perhaps in the diagram, A is equilibrium, but the text says "from A to B and back to A", and in pendulums, usually A and B are extremes.

I think it's correct.

So final answer is 21 m/s².

But let's box it.

Perhaps they want the exact expression, but no, it says "calculate".

So I'll go with 21.

To be precise, let's write it as 21.

But in the calculation, it's 20.56, so maybe 20.6 if three sig figs.

I recall that 0.75 might be considered as having two sig figs, but in some contexts, it's exact.

Perhaps the 1.2 is 1.20, but it's written as 1.2.

I think for this context, since it's a school problem, they might expect 20.6 or 21.

Let's calculate with the values.

Perhaps use g = 4 * 3.14^2 * 0.75 / (1.2)^2

3.14^2 = 9.8596

4*9.8596 = 39.