A particle is executing SHM with amplitude $A$ , time period $T$ , maximum acceleration $a_{0}$ and maximum velocity $v_{0}$ . It starts from mean position at $t = 0$ and at time $t$ it has the displacement $A / 2$ , acceleration $a$ and velocity $v$ , then

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Answer 1

Let the displacement at t is $x = A sin ω t = A sin (\frac{2 π}{T}) t$
here, $x = \frac{A}{2} = A sin (\frac{2 π}{T}) t$
or, $sin (\frac{2 π}{T}) t = \frac{1}{2} = sin \frac{π}{6} \Rightarrow t = \frac{T}{1 2}$
$\frac{d x}{d t} = \frac{2 π}{T} A cos (\frac{2 π}{T}) t$
$\frac{d ^{2} x}{d t ^{2}} = - (\frac{2 π}{T})^{2} A sin (\frac{2 π}{T}) t$
so, $v_{m a x} = v_{0} = (\frac{2 π}{T}) A, a_{m a x} = a_{0} = (\frac{2 π}{T})^{2} A$
now at $t = T / 12,$ the velocity, $v = \frac{d x}{d t} = \frac{2 π}{T} A cos (\frac{2 π}{T}) \frac{T}{1 2}$
$= \frac{2 π}{T} A cos \frac{π}{6} = \frac{2 π}{T} A \frac{3}{2}$
$∴ v = 3 v_{0}$
at $t = T / 12,$ the acceleration, $a = \frac{d ^{2} x}{d t ^{2}} = (\frac{2 π}{T})^{2} A sin (\frac{2 π}{T}) \frac{T}{1 2}$
$= \frac{2 π}{T} A sin \frac{π}{6} = \frac{2 π}{T} A \frac{1}{2}$
$∴ a = \frac{a _{0}}{2}$

Answer 2

Let the displacement at t is

x = A sin ω t = A sin (\frac{2 π}{T}) t

here,

x = \frac{A}{2} = A sin (\frac{2 π}{T}) t

or,

sin (\frac{2 π}{T}) t = \frac{1}{2} = sin \frac{π}{6} \Rightarrow t = \frac{T}{1 2}

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

\frac{d ^{2} x}{d t ^{2}} = - (\frac{2 π}{T})^{2} A sin (\frac{2 π}{T}) t

so,

v_{m a x} = v_{0} = (\frac{2 π}{T}) A, a_{m a x} = a_{0} = (\frac{2 π}{T})^{2} A

now at

t = T / 12,

the velocity,

v = \frac{d x}{d t} = \frac{2 π}{T} A cos (\frac{2 π}{T}) \frac{T}{1 2}

= \frac{2 π}{T} A cos \frac{π}{6} = \frac{2 π}{T} A \frac{3}{2}

∴ v = 3 v_{0}

at

t = T / 12,

the acceleration,

a = \frac{d ^{2} x}{d t ^{2}} = (\frac{2 π}{T})^{2} A sin (\frac{2 π}{T}) \frac{T}{1 2}

= \frac{2 π}{T} A sin \frac{π}{6} = \frac{2 π}{T} A \frac{1}{2}

∴ a = \frac{a _{0}}{2}