Simple harmonic motion

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Question

A particle performs simple harmonic motion with a period of 2 second . The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $\frac{1}{a} s .$ Find the value of $a$ .

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Solution

$L e t s a y x = A \sin (ω t) A t t = 0, x = 0 P u t x = \frac{A}{2} \Rightarrow \frac{A}{2} = A \sin (ω t) \Rightarrow ω t = \frac{π}{6} \Rightarrow \frac{2 π t}{T} = \frac{π}{6} \Rightarrow t = \frac{T}{12}$
Time taken to travel from mean position to half amplitude is $\frac{T}{12}$ , so $\frac{2}{12} = \frac{1}{6} s$

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Simple harmonic motion

Remember concepts with our Masterclasses.

A particle performs simple harmonic motion with a period of 2 second . The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is 1as. Find the value of a.

Let say x=Asinωt At t=0, x=0 Put x= A2 ⇒A2=Asinωt ⇒ωt=π6 ⇒2πtT=π6 ⇒t=T12Time taken to travel from mean position to half amplitude is T12 , so 212=16s

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A particle performs simple harmonic motion with a period of 2 second . The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $\frac{1}{a} s .$ Find the value of $a$ .