The differential equation of a particle executing simple harmonic motion along y-axis is :

Question

The differential equation of a particle executing simple harmonic motion along y-axis is : (A) (d2 y/d t2)+ω2 y=0 (B) (d2 y/d t2)+ω2 y2=0 (C) (d2 y/d t2) - ω2 y=0 (D) (d2 y/d t2)+ω y=0

Tardigrade Admin · Accepted Answer

For a particle executing SHM, restoring force

(F)

is given by

F = ma = m \frac{d v}{d t} = m \frac{d ^{2} y}{d t ^{2}} = - k y

\Rightarrow \frac{d ^{2} y}{d t ^{2}} + \frac{k}{m} y = 0

where

\frac{k}{m} = ω^{2}

is the characteristic angular frequency of the system.
Hence, equation becomes

\frac{d ^{2} y}{d t ^{2}} + ω^{2} y = 0

Q. The differential equation of a particle executing simple harmonic motion along $y$ -axis is :

$\frac{d ^{2} y}{d t ^{2}} + ω^{2} y = 0$

$\frac{d ^{2} y}{d t ^{2}} + ω^{2} y^{2} = 0$

$\frac{d ^{2} y}{d t ^{2}} - ω^{2} y = 0$

$\frac{d ^{2} y}{d t ^{2}} + ω y = 0$

Q. The differential equation of a particle executing simple harmonic motion along y-axis is :

dt2d2y​+ω2y=0

dt2d2y​+ω2y2=0

dt2d2y​−ω2y=0

dt2d2y​+ωy=0

For a particle executing SHM, restoring force (F) is given by F=ma=mdtdv​=mdt2d2y​=−ky ⇒dt2d2y​+mk​y=0 where mk​=ω2 is the characteristic angular frequency of the system. Hence, equation becomes dt2d2y​+ω2y=0

Q. The differential equation of a particle executing simple harmonic motion along $y$ -axis is :

$\frac{d ^{2} y}{d t ^{2}} + ω^{2} y = 0$

$\frac{d ^{2} y}{d t ^{2}} + ω^{2} y^{2} = 0$

$\frac{d ^{2} y}{d t ^{2}} - ω^{2} y = 0$

$\frac{d ^{2} y}{d t ^{2}} + ω y = 0$