A point particle of mass 0.1 kg is executing SHM of amplitude 0.1 m. When the particle passes through the mean position, its kinetic energy is 8 × 10-3 J. The equation of motion of this particle, if its initial phase of oscillation is 45°, is

Question

A point particle of mass 0.1 kg is executing SHM of amplitude 0.1 m. When the particle passes through the mean position, its kinetic energy is 8 × 10-3 J. The equation of motion of this particle, if its initial phase of oscillation is 45°, is (A) y=0.1 sin ((r/4)+(π/4)) (B) y=0.1 sin ((t/2)+(π/4)) (C) y=0.1 sin (4 t-(π/4)) (D) y=0.1 sin (4 t+(π/4))

Tardigrade Admin · Accepted Answer

Kinetic energy at mean position =(1/2) m ω2 a2=8 × 10-3 Or ω=((2 × 8 × 10-3/m a2))1 / 2=[(2 × 8 × 10-3/0.1 ×(0.1)2)]1 / 2=4 Equation of SHM is, y=a sin (ω t+θ)=0.1 sin (4 t+(π/4))

Q. A point particle of mass $0.1 k g$ is executing SHM of amplitude $0.1 m$ . When the particle passes through the mean position, its kinetic energy is $8 \times 1 0^{- 3} J$ . The equation of motion of this particle, if its initial phase of oscillation is $4 5^{\circ}$ , is

$y = 0.1 sin (\frac{r}{4} + \frac{π}{4})$

$y = 0.1 sin (\frac{t}{2} + \frac{π}{4})$

$y = 0.1 sin (4 t - \frac{π}{4})$

$y = 0.1 sin (4 t + \frac{π}{4})$

Kinetic energy at mean position $= \frac{1}{2} m ω^{2} a^{2} = 8 \times 1 0^{- 3}$
Or $ω = (\frac{2 \times 8 \times 1 0 ^{- 3}}{m a ^{2}})^{1/2} = [\frac{2 \times 8 \times 1 0 ^{- 3}}{0.1 \times ( 0.1 ) ^{2}}]^{1/2} = 4$
Equation of SHM is,
$y = a sin (ω t + θ) = 0.1 sin (4 t + \frac{π}{4})$

Q. A point particle of mass 0.1kg is executing SHM of amplitude 0.1m. When the particle passes through the mean position, its kinetic energy is 8×10−3J. The equation of motion of this particle, if its initial phase of oscillation is 45∘, is

y=0.1sin(4r​+4π​)

y=0.1sin(2t​+4π​)

y=0.1sin(4t−4π​)

y=0.1sin(4t+4π​)