A particle is executing S.H.M with the amplitude of 4 cm and period 12 s. The time taken by the particle in going from its mean position to a position of displacement equal to 2 cm is T1. The time taken from this displaced of 2 cm to reach the extreme position is T2. T1/T2 will be.

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Solution

The correct option is A $2$
$\frac{T_{1}}{T_{2}} = \frac{A_{1}}{A_{2}} = \frac{T_{1}}{T_{2}} = \frac{4}{2}$
$A α \frac{1}{f} α T$ or, $\frac{I_{2}}{T_{2}} = 2$
or, $T_{2} = 2 s$ .

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. The time taken by it move from positive extreme position to half the amplitude is:-

A particle executes simple harmonic motion between x = -A and x=+ A. The time taken for it to go from the mean position, O to $\frac{A}{2}$ is $T_{1}$ and to go from $\frac{A}{2}$ to A is $T_{2}$ . Then,

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A particle is executing S.H.M with the amplitude of 4 cm and period 12 s. The time taken by the particle in going from its mean position to a position of displacement equal to 2 cm is T1. The time taken from this displaced of 2 cm to reach the extreme position is T2. T1/T2 will be.

The correct option is A 2T1T2=A1A2=T1T2=42Aα1fαT or, I2T2=2 or, T2=2s.

The correct option is A $2$
$\frac{T_{1}}{T_{2}} = \frac{A_{1}}{A_{2}} = \frac{T_{1}}{T_{2}} = \frac{4}{2}$
$A α \frac{1}{f} α T$ or, $\frac{I_{2}}{T_{2}} = 2$
or, $T_{2} = 2 s$ .