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The kinetic energy of a particle, executing simple harmonic motion is 16 J when it is in its mean position. If the amplitude of oscillations is 25 cm and the mass of the particles is 5.12 kg, the time period of its oscillation is
- a)π/5 sec
- b)2π sec
- c)20π sec
- d)5π sec
Correct answer is option 'A'. Can you explain this answer?
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Given data:
- Kinetic energy at mean position = 16 J
- Amplitude of oscillation = 25 cm = 0.25 m
- Mass of the particle = 5.12 kg
Formula:
- Kinetic energy of a particle in simple harmonic motion is given by:
\(KE = \frac{1}{2} m \omega^2 A^2\)
where,
KE = Kinetic energy
m = Mass of the particle
ω = Angular frequency
A = Amplitude of oscillation
- Time period of oscillation is given by:
\(T = \frac{2\pi}{\omega}\)
Calculation:
Given, KE = 16 J, m = 5.12 kg, A = 0.25 m
Using the formula for kinetic energy, we can find the angular frequency ω:
\(16 = \frac{1}{2} \times 5.12 \times \omega^2 \times (0.25)^2\)
\(16 = 0.32 \times \omega^2 \times 0.0625\)
\(16 = 0.02 \times \omega^2\)
\(\omega^2 = 800\)
\(\omega = 20\)
Now, using the formula for time period, we can find the time period T:
\(T = \frac{2\pi}{20}\)
\(T = \frac{\pi}{10}\)
\(T = \frac{\pi}{5}\) seconds
Therefore, the time period of oscillation is \( \frac{\pi}{5} \) seconds, which corresponds to option A.
- Kinetic energy at mean position = 16 J
- Amplitude of oscillation = 25 cm = 0.25 m
- Mass of the particle = 5.12 kg
Formula:
- Kinetic energy of a particle in simple harmonic motion is given by:
\(KE = \frac{1}{2} m \omega^2 A^2\)
where,
KE = Kinetic energy
m = Mass of the particle
ω = Angular frequency
A = Amplitude of oscillation
- Time period of oscillation is given by:
\(T = \frac{2\pi}{\omega}\)
Calculation:
Given, KE = 16 J, m = 5.12 kg, A = 0.25 m
Using the formula for kinetic energy, we can find the angular frequency ω:
\(16 = \frac{1}{2} \times 5.12 \times \omega^2 \times (0.25)^2\)
\(16 = 0.32 \times \omega^2 \times 0.0625\)
\(16 = 0.02 \times \omega^2\)
\(\omega^2 = 800\)
\(\omega = 20\)
Now, using the formula for time period, we can find the time period T:
\(T = \frac{2\pi}{20}\)
\(T = \frac{\pi}{10}\)
\(T = \frac{\pi}{5}\) seconds
Therefore, the time period of oscillation is \( \frac{\pi}{5} \) seconds, which corresponds to option A.
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The kinetic energy of a particle, executing simple harmonic motion is 16 J when it is in its mean position. If the amplitude of oscillations is 25 cm and the mass of the particles is 5.12 kg, the time period of its oscillation isa)π/5 secb)2π secc)20π secd)5π secCorrect answer is option 'A'. Can you explain this answer?
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