Simple Harmonic Motion
ICSE · Class 11 · Physics
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A simple pendulum has an effective length of 2.5 m. Take g = 10 m/s². What is its time period?
A body in SHM has amplitude 0.20 m and angular frequency 5 rad/s. What is its speed when the displacement is 0.12 m?
A spring-mass system has mass 0.50 kg and force constant 200 N/m. What is its time period?
A simple pendulum is taken from a place where g = 9.8 m/s² to a place where g = 2.45 m/s², with the same length. By what factor does its time period change?
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A pendulum clock on a hill has its effective g reduced from 9.8 m/s² to 8.82 m/s². By what factor does the time period change?
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It increases by a factor of about 1.06
Since T ∝ 1/√g, the new period factor is √(9.8/8.82) = √(1.111...) ≈ 1.054, which is about 1.06. So the period increases slightly.
A spring is cut into two equal parts. If the original spring constant is k, what is the spring constant of each half?
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2k
Spring constant is inversely proportional to length. Cutting the spring into two equal parts halves the length, so each part has twice the spring constant: k' = 2k.
Two springs of constants 3 N/m and 6 N/m are connected in series. What is the equivalent spring constant?
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2 N/m
For series combination, 1/keq = 1/k1 + 1/k2 = 1/3 + 1/6 = 1/2. So keq = 2 N/m.
A simple pendulum has effective length 1.6 m. If its length is increased to 6.4 m, by what factor does the time period change?
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It becomes 2 times
For a simple pendulum, T ∝ √l. The length increases by a factor of 4, so the period increases by √4 = 2.
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