A particle executes simple harmonic motion. Its amplitude is $8 c m$ and time period is $6 s$ . The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is $s$ .

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Solution

$A = 8 c m$

$T = 6 s$

$A cos (\frac{2 π t}{T}) = \frac{A}{2}$

$\Rightarrow \frac{2 π t}{T} = \frac{π}{3}$

or $t = \frac{T}{6} = 1 s$

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A particle executes simple harmonic motion. Its amplitude is 8 cm and time period is 6 s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is s.

A=8 cm T=6 s Acos(2πtT)=A2 ⇒2πtT=π3 or t=T6=1s

A particle executes simple harmonic motion. Its amplitude is $8 c m$ and time period is $6 s$ . The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is $s$ .

$A = 8 c m$

$T = 6 s$

$A cos (\frac{2 π t}{T}) = \frac{A}{2}$

$\Rightarrow \frac{2 π t}{T} = \frac{π}{3}$

or $t = \frac{T}{6} = 1 s$