A particle executing SHM of amplitude 4 cm and T-4s- The time taken by it to move from positive extreme position to half the amplitude is1sdfrac - 1 - 3 - sdfrac - 2 - 3 - ssqrt - dfrac - 3 - 2 - - s

Question

Toppr Admin · Accepted Answer

R-E-F image-xA0-W-2-x3C0-T-2-x3C0-4-x3C0-2 Using Phalov diagram-xA0-wt-x3C0-3-x3C0-2-t-x3C0-3-x21D2-t-23s Ans

A particle executing SHM of amplitude $4 c m$ and $T = 4 s$ . The time taken by it to move from positive extreme position to half the amplitude is
$1 s$
$\frac{1}{3} s$
$\frac{2}{3} s$
$\sqrt{\frac{3}{2}} s$

R.E.F image
$W = \frac{2 π}{T} = \frac{2 π}{4} = \frac{π}{2}$ Using Phalov diagram
$w t = \frac{π}{3}$
$\frac{π}{2} (t) = \frac{π}{3}$
$\Rightarrow t = \frac{2}{3} s$ Ans

A particle executing SHM of amplitude 4cm and T=4s. The time taken by it to move from positive extreme position to half the amplitude is1s13s23s√32s

R.E.F image W=2πT=2π4=π2 Using Phalov diagram wt=π3π2(t)=π3⇒t=23s Ans

A particle executing SHM of amplitude $4 c m$ and $T = 4 s$ . The time taken by it to move from positive extreme position to half the amplitude is
$1 s$
$\frac{1}{3} s$
$\frac{2}{3} s$
$\sqrt{\frac{3}{2}} s$

R.E.F image
$W = \frac{2 π}{T} = \frac{2 π}{4} = \frac{π}{2}$ Using Phalov diagram
$w t = \frac{π}{3}$
$\frac{π}{2} (t) = \frac{π}{3}$
$\Rightarrow t = \frac{2}{3} s$ Ans