Question 131: Oscillations

Question

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is -

Accepted Answer

$$v = \omega \sqrt {{A^2} - {x^2}} \,\,$$    . . .(1)
$$a =  - {\omega ^2}x$$               . . .(2)
$$\left| v ight| = \left| a ight|$$                   . . .(3)
$$\omega \sqrt {{A^2} - {x^2}}  = {\omega ^2}x$$
$${A^2} - {x^2} = {\omega ^2}{x^2}$$
$${5^2} - {4^2} = {\omega ^2}\left( {{4^2}} ight)$$
$$ \Rightarrow \,\,\,3 = \omega  	imes 4$$
$$T = 2\pi /\omega $$

Answer

$${{4\pi } \over 3}$$

Answer

$${3 \over 8}\pi $$

Answer

$${7 \over 3}\pi $$

Answer

$${{8\pi } \over 3}$$

Oscillations

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is -

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