A particle is oscillating according to the equation X=7 cos 0.5 π t, where ' t ' is in second. The point moves from the position of equilibrium to maximum displacement in time (in sec)?

Question

A particle is oscillating according to the equation X=7 cos 0.5 π t, where ' t ' is in second. The point moves from the position of equilibrium to maximum displacement in time (in sec)? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

x=7 cos 0.5 π General form of equation. x=A cos ω t Given equation compare with standard form ω=(2 π/T)=0.5 π ; T =4 sec T'= T / 4=4 / 4=1 sec

Q. A particle is oscillating according to the equation $X = 7 cos 0.5 π t$ , where ' $t$ ' is in second. The point moves from the position of equilibrium to maximum displacement in time (in $sec$ )?

$x = 7 cos 0.5 π$
General form of equation.
$x = A cos ω t$
Given equation compare with standard form
$ω = \frac{2 π}{T} = 0.5 π; T = 4 sec$
$T^{'} = T /4 = 4/4 = 1 sec$

Q. A particle is oscillating according to the equation X=7cos0.5πt, where ' t ' is in second. The point moves from the position of equilibrium to maximum displacement in time (in sec)?

x=7cos0.5π General form of equation. x=Acosωt Given equation compare with standard form ω=T2π​=0.5π;T=4sec T′=T/4=4/4=1sec

Q. A particle is oscillating according to the equation $X = 7 cos 0.5 π t$ , where ' $t$ ' is in second. The point moves from the position of equilibrium to maximum displacement in time (in $sec$ )?

$x = 7 cos 0.5 π$
General form of equation.
$x = A cos ω t$
Given equation compare with standard form
$ω = \frac{2 π}{T} = 0.5 π; T = 4 sec$
$T^{'} = T /4 = 4/4 = 1 sec$