QUARTER FINAL STAGE 2024
YILO KROBO
KOFORIDUA SENIOR HIGH TECH
KNUST SHS
ROUND 1
SECOND SET OF QUESTIONS
PREAMBLE
QUESTION
Find the length of a simple pendulum on earth with the given property (answers in 2 significant figures).
1. Its period is 2.5 seconds
ANSWER: $L = 1.6 \text{ m}$
FORMULA
$L \approx \dfrac{T^2}{4}$ (using $\dfrac{g}{\pi^2} \approx 1$)
SOLUTION:
$T = 2.5$
$T = \dfrac{5}{2}$
$T^2 = \dfrac{25}{4}$
$T^2 = 6.25$
$L \approx \dfrac{6.25}{4}$
$L = 1.56 \text{ m}$
$L \approx 1.6 \text{ m}$
ANSWER: $L = 1.6 \text{ m}$
2. Its frequency is 0.42 Hertz
ANSWER: $L = 1.4 \text{ m}$
FORMULA
$L \approx \dfrac{1}{4f^2}$ (using $T = \dfrac{1}{f}$ and $\dfrac{g}{\pi^2} \approx 1$)
SOLUTION:
$f = 0.42$
$f \approx \dfrac{5}{12}$
$f^2 \approx \dfrac{25}{144}$
$L \approx \dfrac{1}{4 \times \dfrac{25}{144}}$
$L = \dfrac{144}{100}$
$L = \dfrac{36}{25}$
$L = 1.44 \text{ m}$
$L \approx 1.4 \text{ m}$
ANSWER: $L = 1.4 \text{ m}$
3. Its angular frequency is 3.2 rad/s
ANSWER: $L = 0.96 \text{ m}$
FORMULA
$L = \dfrac{g}{\omega^2}$
SOLUTION:
$\omega = 3.2$
$\omega = \dfrac{16}{5}$
$\omega^2 = \dfrac{256}{25}$
$\omega^2 = 10.24$
$L = \dfrac{9.8}{10.24}$
$L \approx 0.957 \text{ m}$
$L \approx 0.96 \text{ m}$
ANSWER: $L = 0.96 \text{ m}$
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PRACTICE QUESTIONS
1. QUESTION: Find the length of a simple pendulum on Earth whose period is 2.4 s.
ANSWER: 1.4 m
SOLUTION:
$T = 2\pi\sqrt{\dfrac{L}{g}}$
$L = \dfrac{gT^2}{4\pi^2}$
$\dfrac{g}{\pi^2} \approx 1$ (9.8 ÷ 9.87)
$L \approx \dfrac{T^2}{4}$
$L = \dfrac{2.4 \times 2.4}{4}$
$L = \dfrac{24\times10^{-1} \times 24\times10^{-1}}{4}$
$L = 6 \times 24 \times10^{-2}$
$L \approx 1.4\text{ m}$
2. QUESTION: Find the length of a simple pendulum on Earth whose period is 1.2 s.
ANSWER: 0.36 m or $3.6\times10^{-1}$ m
SOLUTION:
$T = 2\pi\sqrt{\dfrac{L}{g}}$
$L = \dfrac{gT^2}{4\pi^2}$
$\dfrac{g}{\pi^2} \approx 1$ (9.8 ÷ 9.87)
$L \approx \dfrac{T^2}{4}$
$L = \dfrac{1.2 \times 1.2}{4}$
$L = \dfrac{12\times10^{-1} \times 12\times10^{-1}}{4}$
$L = 3 \times 12 \times10^{-2}$
$L = 0.36\text{ m}$
3. QUESTION: Find the length of a simple pendulum on Earth whose period is 1.4 s.
ANSWER: 0.49 m or $4.9\times10^{-1}$ m
SOLUTION:
$T = 2\pi\sqrt{\dfrac{L}{g}}$
$L = \dfrac{gT^2}{4\pi^2}$
$\dfrac{g}{\pi^2} \approx 1$ (9.8 ÷ 9.87)
$L \approx \dfrac{T^2}{4}$
$L = \dfrac{1.4 \times 1.4}{4}$
$L = \dfrac{14\times10^{-1} \times 14\times10^{-1}}{4}$
$L = \dfrac{7 \times 14}{2} \times10^{-2}$
$L = 7 \times 7 \times10^{-2}$
$L = 0.49\text{ m}$
4. QUESTION: Find the length of a simple pendulum on Earth whose period is 1.6 s.
ANSWER: 0.64 m or $6.4\times10^{-1}$ m
SOLUTION:
$T = 2\pi\sqrt{\dfrac{L}{g}}$
$L = \dfrac{gT^2}{4\pi^2}$
$\dfrac{g}{\pi^2} \approx 1$ (9.8 ÷ 9.87)
$L \approx \dfrac{T^2}{4}$
$L = \dfrac{1.6 \times 1.6}{4}$
$L = \dfrac{16\times10^{-1} \times 16\times10^{-1}}{4}$
$L = 4 \times 16 \times10^{-2}$
$L = 0.64\text{ m}$
5. QUESTION: Find the length of a simple pendulum on Earth whose frequency is 0.25 Hz.
ANSWER: 4.0 m
SOLUTION:
$T = 2\pi\sqrt{\dfrac{L}{g}}$
$L = \dfrac{gT^2}{4\pi^2}$
$\dfrac{g}{\pi^2} \approx 1$ (9.8 ÷ 9.87)
$L \approx \dfrac{1}{4f^2}$ (with $T = \dfrac{1}{f}$)
$L = \dfrac{1}{4 \times 0.25 \times 0.25}$
$L = \dfrac{1}{4 \times 25\times10^{-2} \times 25\times10^{-2}}$
$L = \dfrac{1}{4 \times 25 \times 25} \times10^{4}$
$L = 4.0\text{ m}$
6. QUESTION: Find the length of a simple pendulum on Earth whose frequency is 2.5 Hz.
ANSWER: 0.040 m or $4.0\times10^{-2}$ m
SOLUTION:
$T = 2\pi\sqrt{\dfrac{L}{g}}$
$L = \dfrac{gT^2}{4\pi^2}$
$\dfrac{g}{\pi^2} \approx 1$ (9.8 ÷ 9.87)
$L \approx \dfrac{1}{4f^2}$ (with $T = \dfrac{1}{f}$)
$L = \dfrac{1}{4 \times 2.5 \times 2.5}$
$L = \dfrac{1}{4 \times 25\times10^{-1} \times 25\times10^{-1}}$
$L = \dfrac{1}{4 \times 25 \times 25} \times10^{2}$
$L = 0.040\text{ m}$
7. QUESTION: Find the length of a simple pendulum on Earth whose angular frequency is 7.0 rad/s.
ANSWER: 0.20 m or $2.0\times10^{-1}$ m
SOLUTION:
$\omega = \sqrt{\dfrac{g}{L}}$
$L = \dfrac{g}{\omega^2}$
$L = \dfrac{9.8}{7.0 \times 7.0}$
$L = \dfrac{98\times10^{-1}}{7 \times 7}$
$L = \dfrac{14}{7} \times10^{-1}$
$L = 0.20\text{ m}$
8. QUESTION: Find the length of a simple pendulum on Earth whose angular frequency is 3.5 rad/s.
ANSWER: 0.80 m or $8.0\times10^{-1}$ m
SOLUTION:
$\omega = \sqrt{\dfrac{g}{L}}$
$L = \dfrac{g}{\omega^2}$
$L = \dfrac{9.8}{3.5 \times 3.5}$
$L = \dfrac{98\times10^{-1}}{35\times10^{-1} \times 35\times10^{-1}}$
$L = \dfrac{14}{5 \times 35} \times10^{1}$
$L = \dfrac{2}{5 \times 5} \times10^{1}$
$L = 0.80\text{ m}$
9. QUESTION: Find the length of a simple pendulum on Earth whose angular frequency is 1.4 rad/s.
ANSWER: 5.0 m
SOLUTION:
$\omega = \sqrt{\dfrac{g}{L}}$
$L = \dfrac{g}{\omega^2}$
$L = \dfrac{9.8}{1.4 \times 1.4}$
$L = \dfrac{98\times10^{-1}}{14\times10^{-1} \times 14\times10^{-1}}$
$L = \dfrac{7}{14} \times10^{1}$
$L = \dfrac{1}{2} \times10^{1}$
$L = 5.0\text{ m}$