TS EAMCET 4-Aug-2021 Shift 2 Question Paper

Distance of particle from mean position,
x1=1cm
Acceleration, a=3cm∕s2
Velocity, v2=6cm∕s, when x2=2cm.
As we know that, v2=ω√A2−x22
∴‌‌6=ω√A2−22
On squaring both sides, we get
62=ω2(A2−4) . . . (i)
and since, ‌‌a=−ω2x
where, a is acceleration.
∴‌‌3=ω2
⇒‌‌ω2=3 put in Eq. (i), we get
36=3(A2−4)
⇒‌‌12=A2−4
⇒‌‌A2=12+4=16
⇒‌‌A=4cm