A particle executing SHM according to the equation x= 5 cos (2π t +(π/4)) in SI units. The displacement and acceleration of the particle at t = 1.5 s is

Question

A particle executing SHM according to the equation x= 5 cos (2π t +(π/4)) in SI units. The displacement and acceleration of the particle at t = 1.5 s is (A) -3.0 m, 100 m s-2 (B) +2.54 m, 200 m s-2 (C) -3.54 m, 140 m s-2 (D) +3.55 m, 120 m s-2

Tardigrade Admin · Accepted Answer

The given equation of simple harmonic motion is

x (t) = 5 cos (2 π t + \frac{π}{4})

Compare the given equation with standard equation of

S H M

x (t) = A cos (ω t + ϕ)

we get,

ω = 2 π s^{- 1}

At

t = 1.5 s

Displacement,

x (t) = 5 cos (2 π \times 1.5 + \frac{π}{4})

= 5 cos (3 π + \frac{π}{4}) = - 5 cos (\frac{π}{4})

[∵ cos (3 π + θ) = - cos θ]

= - 5 \times 0.707 m

= - 3.54 m

Accelaration,

a = - ω^{2} \times

displacement

= - (2 π s^{- 1})^{2} \times (- 3.54 m)

= 140 m s^{- 1}

Q. A particle executing $S H M$ according to the equation
$x = 5 cos (2 π t + \frac{π}{4})$ in $S I$ units. The displacement and acceleration of the particle at $t = 1.5 s$ is

$- 3.0 m, 100 m s^{-} 2$

$+ 2.54 m, 200 m s^{-} 2$

$- 3.54 m, 140 m s^{-} 2$

$+ 3.55 m, 120 m s^{-} 2$

Q. A particle executing SHM according to the equation x=5cos(2πt+4π​) in SI units. The displacement and acceleration of the particle at t=1.5s is

−3.0m,100ms−2

+2.54m,200ms−2

−3.54m,140ms−2

+3.55m,120ms−2

Q. A particle executing $S H M$ according to the equation
$x = 5 cos (2 π t + \frac{π}{4})$ in $S I$ units. The displacement and acceleration of the particle at $t = 1.5 s$ is

$- 3.0 m, 100 m s^{-} 2$

$+ 2.54 m, 200 m s^{-} 2$

$- 3.54 m, 140 m s^{-} 2$

$+ 3.55 m, 120 m s^{-} 2$