The equation describing the motion of a simple harmonic oscillator along the x axis is given as x=A cos (ω t+φ). If at time t=0, the oscillator is at x=0 and moving in the negative x-direction, then the phase angle φ is

Question

The equation describing the motion of a simple harmonic oscillator along the x axis is given as x=A cos (ω t+φ). If at time t=0, the oscillator is at x=0 and moving in the negative x-direction, then the phase angle φ is (A)  (π /2)  (B)  -(π /2)  (C)  π  (D)  0

Tardigrade Admin · Accepted Answer

x=A cos (ω t+φ) Let t=0 x=0 0= cos φ cos (π/2)= cos φ θ=(π/2)

Q. The equation describing the motion of a simple harmonic oscillator along the $x$ axis is given as $x = A cos (ω t + ϕ)$ . If at time $t = 0$ , the oscillator is at $x = 0$ and moving in the negative $x$ -direction, then the phase angle $ϕ$ is

$\frac{π}{2}$

$- \frac{π}{2}$

$π$

$0$

$x = A cos (ω t + ϕ)$
Let $t = 0$
$x = 0$
$0 = cos ϕ$
$cos \frac{π}{2} = cos ϕ$
$θ = \frac{π}{2}$

Q. The equation describing the motion of a simple harmonic oscillator along the x axis is given as x=Acos(ωt+ϕ). If at time t=0, the oscillator is at x=0 and moving in the negative x-direction, then the phase angle ϕ is

2π​

−2π​

π

0

x=Acos(ωt+ϕ) Let t=0 x=0 0=cosϕ cos2π​=cosϕ θ=2π​

Q. The equation describing the motion of a simple harmonic oscillator along the $x$ axis is given as $x = A cos (ω t + ϕ)$ . If at time $t = 0$ , the oscillator is at $x = 0$ and moving in the negative $x$ -direction, then the phase angle $ϕ$ is

$\frac{π}{2}$

$- \frac{π}{2}$

$π$

$0$

$x = A cos (ω t + ϕ)$
Let $t = 0$
$x = 0$
$0 = cos ϕ$
$cos \frac{π}{2} = cos ϕ$
$θ = \frac{π}{2}$