Displaystyle y-2left- cm right - sin left - frac-pi t-2- - phi right - what is the maximum acceleration of the particle doing the S-H-M -displaystyle dfrac-pi-2- cm-s-2-displaystyle dfrac-pi-2-2- cm-s-2-displaystyle dfrac-pi-4- cm-s-2-displaystyle dfrac-pi-2-4- cm-s-2-

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Toppr Admin · Accepted Answer

Y-2sin-x3C0-t2-x3D5-Velocity of particle dydt-2-xD7-x3C0-2cos-x3C0-t2-x3D5-Acceleration-xA0-d2ydt-x2212-x3C0-242sin-x3C0-t2-x3D5-Thus amax-x3C0-22-xA0-

$y = 2 (c m) sin [\frac{π t}{2} + ϕ]$ what is the maximum acceleration of the particle doing the S.H.M ?
$\frac{π}{2} c m / s^{2}$
$\frac{π^{2}}{2} c m / s^{2}$
$\frac{π}{4} c m / s^{2}$
$\frac{π^{2}}{4} c m / s^{2}$

$y = 2 sin (\frac{π t}{2} + ϕ)$
Velocity of particle $\frac{d y}{d t} = 2 \times \frac{π}{2} cos (\frac{π t}{2} + ϕ)$
Acceleration $\frac{d^{2} y}{d t} = - \frac{π^{2}}{4} 2 sin (\frac{π t}{2} + ϕ)$
Thus $a_{m a x} = \frac{π^{2}}{2}$

y=2(cm)sin[πt2+ϕ] what is the maximum acceleration of the particle doing the S.H.M ?π2 cm/s2π22 cm/s2π4 cm/s2π24 cm/s2

y=2sin(πt2+ϕ)Velocity of particle dydt=2×π2cos(πt2+ϕ)Acceleration d2ydt=−π242sin(πt2+ϕ)Thus amax=π22

$y = 2 (c m) sin [\frac{π t}{2} + ϕ]$ what is the maximum acceleration of the particle doing the S.H.M ?
$\frac{π}{2} c m / s^{2}$
$\frac{π^{2}}{2} c m / s^{2}$
$\frac{π}{4} c m / s^{2}$
$\frac{π^{2}}{4} c m / s^{2}$

$y = 2 sin (\frac{π t}{2} + ϕ)$
Velocity of particle $\frac{d y}{d t} = 2 \times \frac{π}{2} cos (\frac{π t}{2} + ϕ)$
Acceleration $\frac{d^{2} y}{d t} = - \frac{π^{2}}{4} 2 sin (\frac{π t}{2} + ϕ)$
Thus $a_{m a x} = \frac{π^{2}}{2}$