SATHEE: Oscillations - Result Question 35

Oscillations - Result Question 35

39. A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 $m / s^{2}$ at a distance of $5 m$ from the mean position. The time period of oscillation is

(a) $2 π s$

(b) $π s$

(c) $1 s$

(d) $2 s$

[2018]

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Answer:

Correct Answer: 39. (b)

Solution:

(b) From question, acceleration, $a = 20 m / s^{2}$ , and displacement, $y = 5 m$

$| a | = ω^{2} y$

$\Rightarrow 20 = ω^{2} (5)$

$\Rightarrow ω = 2 r a d / s$

Time period of pendulum,

$T = \frac{2 π}{ω} = \frac{2 π}{2} = π s$

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Oscillations - Result Question 35

39. A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s2 at a distance of 5m from the mean position. The time period of oscillation is

Answer:

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39. A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 $m / s^{2}$ at a distance of $5 m$ from the mean position. The time period of oscillation is