A particle executes simple harmonic motion (amplitude = A) between x =

Q: A particle executes simple harmonic motion (amplitude $= A$) between $x = - A$ and $x = + A$. The time taken for it to go from $0$ to $A/2$ is ${T_1}$ and to go from $A/2$ to $A$ is ${T_2}$. Then

a (a)Using $x = A\sin \omega t$ For $x = A/2,\;\;\sin \omega {T_1} = 1/2 \Rightarrow {T_1} = \frac{\pi }{{6\omega }}$ For $x = A,\;\sin \omega ({T_1} + {T_2}) = 1 \Rightarrow {T_1} + {T_2} = \frac{\pi }{{2\omega }}$ $ \Rightarrow {T_2} = \frac{\pi }{{2\omega }} - {T_1} = \frac{\pi }{{2\omega }} - \frac{\pi }{{6\omega }} = \frac{\pi }{{3\omega }} $ $i.e.\;{T_1} < {T_2}$ Alternate method : In S.H.M., velocity of particle also oscillates simple harmonically. Speed is more near the mean position and less near the extreme position. Therefore the time taken for the particle to go from $0$ to $\frac{A}{2}$ will be less than the time taken to go from $\frac{A}{2}$ to $A$. Hence ${T_1} < {T_2}.$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

A particle executes simple harmonic motion (amplitude $= A$) between $x = - A$ and $x = + A$. The time taken for it to go from $0$ to $A/2$ is ${T_1}$ and to go from $A/2$ to $A$ is ${T_2}$. Then

A${T_1} < {T_2}$
B${T_1} > {T_2}$
C${T_1} = {T_2}$
D${T_1} = 2{T_2}$

IIT 2001, Medium

STD 11 Science
NEET
STD 11 - 13. oscillations
Physics

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