The displacement of a particle executing simple harmonic motion is given by x = 3 sin (2π t+(π/4)) where x is in metres and t is in seconds. The amplitude and maximum speed of the particle is

Question

The displacement of a particle executing simple harmonic motion is given by x = 3 sin (2π t+(π/4)) where x is in metres and t is in seconds. The amplitude and maximum speed of the particle is (A) 3 m, 2π m s-1 (B) 3 m, 4π m s-1 (C) 3 m, 6π m s-1 (D) 3 m, 8π m s-1

Tardigrade Admin · Accepted Answer

The given equation of SHM is x = 3 sin (2π t+(π/4)) Compare the given equation with standard equation of SHM x = A sin (ω t+φ) we get, A = 3 m, ω = 2π s-1 ∴ Maximum speed, v textmax = Aω = 3 m × 2 π s-1 = 6 π m s-1

Q. The displacement of a particle executing simple harmonic motion is given by
$x = 3 s in (2 π t + \frac{π}{4})$
where $x$ is in metres and $t$ is in seconds. The amplitude and maximum speed of the particle is

$3 m, 2 π m s^{- 1}$

$3 m, 4 π m s^{- 1}$

$3 m, 6 π m s^{- 1}$

$3 m, 8 π m s^{- 1}$

The given equation of $S H M$ is
$x = 3 s in (2 π t + \frac{π}{4})$
Compare the given equation with standard equation of $S H M$
$x = A s in (ω t + ϕ)$
we get, $A = 3 m, ω = 2 π s^{- 1}$
$∴$ Maximum speed, $v_{max} = A ω = 3 m \times 2 π s^{- 1}$
$= 6 π m s^{- 1}$

Q. The displacement of a particle executing simple harmonic motion is given by x=3sin(2πt+4π​) where x is in metres and t is in seconds. The amplitude and maximum speed of the particle is

3m,2πms−1

3m,4πms−1

3m,6πms−1

3m,8πms−1