The displacement y(t)=A sin(ω t+φ) of a pendulum for φ=(2π/3) is correctly represented by

Question

The displacement y(t)=A sin(ω t+φ) of a pendulum for φ=(2π/3) is correctly represented by (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Displacement y(t)=A sin (wt+φ) [Given] For φ=(2π/3) at t=0; y=A sin φ=A sin (2π/3) =A sin 120°=0.87 A[∵ sin 120°≃0.866] Graph (a) depicts y = 0.87 at t=0

Q. The displacement $y (t) = A s in (ω t + ϕ)$ of a pendulum for $ϕ = \frac{2 π}{3}$ is correctly represented by

Displacement $y (t) = A s in (wt + ϕ)$
[Given]
For $ϕ = \frac{2 π}{3} < b r / > a tt = 0; y = A s in ϕ = A s in \frac{2 π}{3}$
$= A s in 12 0^{\circ} = 0.87 A [∵ s in 12 0^{\circ} ≃ 0.866]$
Graph (a) depicts $y = 0.87$ at $t = 0$

Q. The displacement y(t)=Asin(ωt+ϕ) of a pendulum for ϕ=32π​ is correctly represented by

Displacement y(t)=Asin(wt+ϕ) [Given] For ϕ=32π​<br/>att=0;y=Asinϕ=Asin32π​ =Asin120∘=0.87A[∵sin120∘≃0.866] Graph (a) depicts y=0.87 at t=0

Q. The displacement $y (t) = A s in (ω t + ϕ)$ of a pendulum for $ϕ = \frac{2 π}{3}$ is correctly represented by

Displacement $y (t) = A s in (wt + ϕ)$
[Given]
For $ϕ = \frac{2 π}{3} < b r / > a tt = 0; y = A s in ϕ = A s in \frac{2 π}{3}$
$= A s in 12 0^{\circ} = 0.87 A [∵ s in 12 0^{\circ} ≃ 0.866]$
Graph (a) depicts $y = 0.87$ at $t = 0$