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Pendulum worksheet

Grades 9–10 · CCSS HSA.CED.A.4

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Find: Length for a period
- Period (T): 3 s
- Show length in: Meters (m)
- Gravity (g): Mars: 3.71 m/s²
Find: Pendulum length
Answer
2.
Problem 2.
- Find: Length for a period
- Period (T): 2 s
- Show length in: Millimeters (mm)
- Gravity (g): Custom value
- Custom gravitational acceleration: 7.8 m/s²
Find: Pendulum length
Answer
3.
Problem 3.
- Find: Period from length
- Pendulum length (L): 1 m
- Gravity (g): Custom value
- Custom gravitational acceleration: 10.5 m/s²
Find: Period (small swings)
Answer
4.
Problem 4.
- Find: Length for a period
- Period (T): 2 s
- Show length in: Millimeters (mm)
- Gravity (g): Custom value
- Custom gravitational acceleration: 11.75 m/s²
Find: Pendulum length
Answer
5.
Problem 5.
- Find: Period from length
- Pendulum length (L): 1 m
- Gravity (g): Moon: 1.62 m/s²
Find: Period (small swings)
Answer
6.
Problem 6.
- Find: Length for a period
- Period (T): 3 s
- Show length in: Feet (ft)
- Gravity (g): Moon: 1.62 m/s²
Find: Pendulum length
Answer
7.
Problem 7.
- Find: Length for a period
- Period (T): 2 s
- Show length in: Inches (in)
- Gravity (g): Custom value
- Custom gravitational acceleration: 11 m/s²
Find: Pendulum length
Answer
8.
Problem 8.
- Find: Period from length
- Pendulum length (L): 1 m
- Gravity (g): Earth: 9.80665 m/s² (standard)
Find: Period (small swings)
Answer
9.
Problem 9.
- Find: Length for a period
- Period (T): 2 s
- Show length in: Meters (m)
- Gravity (g): Custom value
- Custom gravitational acceleration: 9 m/s²
Find: Pendulum length
Answer
10.
Problem 10.
- Find: Length for a period
- Period (T): 1 s
- Show length in: Centimeters (cm)
- Gravity (g): Earth: 9.80665 m/s² (standard)
Find: Pendulum length
Answer

Answer key: Pendulum worksheet

1. Pendulum length: 0.85 m
1. Use the surface gravity of Mars
  $g$ $= 3.71 m/s²$
2. Turn the period formula around to solve for the length L: gravity g × (period T ÷ 2π)²
  $L$ $= 3.71 \times {(\frac{3}{2 \times π})}^{2}$ $= 0.8457785805 m$
2. Pendulum length: 790.31 mm
1. Use the gravity value you entered
  $g$ $= 7.8 m/s²$
2. Turn the period formula around to solve for the length L: gravity g × (period T ÷ 2π)²
  $L$ $= 7.8 \times {(\frac{2}{2 \times π})}^{2}$ $= 0.7903052324 m$
3. Change meters to mm: divide by 0.001
  $\frac{0.7903052324}{0.001}$ $= 790.3052324102 mm$
3. Period (small swings): 1.94 s
1. Use the gravity value you entered
  $g$ $= 10.5 m/s²$
2. Find the small-swing period T₀ from the length L and gravity g: 2π × the square root of L ÷ g
  $T₀$ $= 2 \times π \times \sqrt{\frac{1}{10.5}}$ $= 1.9390330827 s$
3. Frequency f is the number of full swings per second, so it's 1 ÷ the period T
  $f$ $= \frac{1}{1.9390330827}$ $= 0.5157209585 Hz$
4. Pendulum length: 1,190.52 mm
1. Use the gravity value you entered
  $g$ $= 11.75 m/s²$
2. Turn the period formula around to solve for the length L: gravity g × (period T ÷ 2π)²
  $L$ $= 11.75 \times {(\frac{2}{2 \times π})}^{2}$ $= 1.1905239078 m$
3. Change meters to mm: divide by 0.001
  $\frac{1.1905239078}{0.001}$ $= 1,190.5239077975 mm$
5. Period (small swings): 4.94 s
1. Use the surface gravity of the Moon
  $g$ $= 1.62 m/s²$
2. Find the small-swing period T₀ from the length L and gravity g: 2π × the square root of L ÷ g
  $T₀$ $= 2 \times π \times \sqrt{\frac{1}{1.62}}$ $= 4.936536598 s$
3. Frequency f is the number of full swings per second, so it's 1 ÷ the period T
  $f$ $= \frac{1}{4.936536598}$ $= 0.2025711711 Hz$
6. Pendulum length: 1.21 ft
1. Use the surface gravity of the Moon
  $g$ $= 1.62 m/s²$
2. Turn the period formula around to solve for the length L: gravity g × (period T ÷ 2π)²
  $L$ $= 1.62 \times {(\frac{3}{2 \times π})}^{2}$ $= 0.3693157144 m$
3. Change meters to ft: divide by 0.3048
  $\frac{0.3693157144}{0.3048}$ $= 1.2116657296 ft$
7. Pendulum length: 43.88 in
1. Use the gravity value you entered
  $g$ $= 11 m/s²$
2. Turn the period formula around to solve for the length L: gravity g × (period T ÷ 2π)²
  $L$ $= 11 \times {(\frac{2}{2 \times π})}^{2}$ $= 1.1145330201 m$
3. Change meters to in: divide by 0.0254
  $\frac{1.1145330201}{0.0254}$ $= 43.8792527585 in$
8. Period (small swings): 2.01 s
1. Use the surface gravity of Earth (standard gravity)
  $g$ $= 9.80665 m/s²$
2. Find the small-swing period T₀ from the length L and gravity g: 2π × the square root of L ÷ g
  $T₀$ $= 2 \times π \times \sqrt{\frac{1}{9.80665}}$ $= 2.0064092926 s$
3. Frequency f is the number of full swings per second, so it's 1 ÷ the period T
  $f$ $= \frac{1}{2.0064092926}$ $= 0.4984027953 Hz$
9. Pendulum length: 0.91 m
1. Use the gravity value you entered
  $g$ $= 9 m/s²$
2. Turn the period formula around to solve for the length L: gravity g × (period T ÷ 2π)²
  $L$ $= 9 \times {(\frac{2}{2 \times π})}^{2}$ $= 0.9118906528 m$
10. Pendulum length: 24.84 cm
1. Use the surface gravity of Earth (standard gravity)
  $g$ $= 9.80665 m/s²$
2. Turn the period formula around to solve for the length L: gravity g × (period T ÷ 2π)²
  $L$ $= 9.80665 \times {(\frac{1}{2 \times π})}^{2}$ $= 0.2484053464 m$
3. Change meters to cm: divide by 0.01
  $\frac{0.2484053464}{0.01}$ $= 24.8405346392 cm$