If the magnitude of displacement is numerically equal to that of acceleration, then the time period is:
$π s e c o n d s$
$1 s e c o n d$
$2 π s e c o n d s$
$4 π s e c o n d s$

Question

If the magnitude of displacement is numerically equal to that of acceleration- then the time period is-pispace seconds1space second2pispace seconds4pispace seconds

Toppr Admin · Accepted Answer

-xA8-y-x3C9-2y-x21D2-x3C9-1-x21D2-4-x3C0-2T2-1 -x21D2-T2-4-x3C0-2-x21D2-T-2-x3C0-xA0-seconds

If the magnitude of displacement is numerically equal to that of acceleration, then the time period is:π seconds1 second2π seconds4π seconds

¨y=ω2y⇒ω=1⇒4π2T2=1 ⇒T2=4π2⇒T=2π seconds

If the magnitude of displacement is numerically equal to that of acceleration, then the time period is:
$π s e c o n d s$
$1 s e c o n d$
$2 π s e c o n d s$
$4 π s e c o n d s$

$¨ y = ω^{2} y \Rightarrow ω = 1 \Rightarrow \frac{4 π^{2}}{T^{2}} = 1$
$\Rightarrow T^{2} = 4 π^{2}$
$\Rightarrow T = 2 π s e c o n d s$