SATHEE: Oscillations - Result Question 45

Oscillations - Result Question 45

49. A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is $T$ . The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

(a) $T / 8$

(b) $T / 12$

(c) $T / 2$

(d) $T / 4$

[2007]

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Answer:

Correct Answer: 49. (b)

Solution:

(b) Displacement from the mean position

$y = a \sin (\frac{2 π}{T}) t$

According to problem $y = a / 2$

$a / 2 = a \sin (\frac{2 π}{T}) t$

$\Rightarrow \frac{π}{6} = (\frac{2 π}{T}) t \Rightarrow t = T / 12$

This is the minimum time taken by the particle to travel half of the amplitude from the equilibrium position.

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Oscillations - Result Question 45

49. A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

Answer:

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49. A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is $T$ . The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is