A particle is executing S.H.M. with amplitude A and has maximum velocity v0. Its speed at displacement (3 A/4) will be

Question

A particle is executing S.H.M. with amplitude A and has maximum velocity v0. Its speed at displacement (3 A/4) will be (A) (√7/4) v0 (B) (v0/√2) (C) v0 (D) (√3/2) v0

Tardigrade Admin · Accepted Answer

v=ω √A2-x2 x=(3/4) A v=ω √A2-(9 A2/16) v=ω A √(7/16) or v=v0 √(7/4) as (v0=A ω)

Q. A particle is executing S.H.M. with amplitude $A$ and has maximum velocity $v_{0}$ . Its speed at displacement $\frac{3 A}{4}$ will be

$\frac{7}{4} v_{0}$

$\frac{v _{0}}{2}$

$v_{0}$

$\frac{3}{2} v_{0}$

$v = ω A^{2} - x^{2}$
$x = \frac{3}{4} A$
$v = ω A^{2} - \frac{9 A ^{2}}{16}$
$v = ω A \frac{7}{16}$
or $v = v_{0} \frac{7}{4}$ as $(v_{0} = A ω)$

Q. A particle is executing S.H.M. with amplitude A and has maximum velocity v0​. Its speed at displacement 43A​ will be

47​​v0​

2​v0​​

v0​

23​​v0​