Verified answer

The displacement of the bob in radians is given by the equation

(1) θ(t) = (0.50 rad) cos(3t)

The maximum and minimum values of the cos function are +1 and -1; therefore, the amplitude A--the max distance from the equilibrium point--is:

(2) A = 0.5rad

Equation (1) has the general form:

(3) θ(t) = A * cos(ω * t),

where ω = angular velocity = 3 rad/s = 2 * π * f; so

(3.1) f = ω / 2 * π = 3 / (2 * 3.14) = 0.478Hz

The length of a pendulum string L is related to the period by:

(4) T = 2 * π * sqrt (L / g) . . . . (see ref. 1)

Since T = 1 / f and f = ω / 2 * π ,

(5) 2 * π / ω = 2 * π * sqrt (L / g); canceling the 2 * π we get:

(6) 1 / ω = sqrt (L / g); . . . . squaring, we get:

(7) 1 / ω^2 = L / g; . . . . . solving for L:

(8) g / ω^2 = L

= 9.81 / 3^2

= 1.09 m . . . . . . . .<<== length of string

To get the angle at 2 sec, use (1):

(9) θ(2) = (0.50 rad) cos(3 * 2)

= 0.480 rad

(a) The amplitude.

____0.5____rad

(b) The frequency.

_____0.478____Hz

(c) The length of the string.

____1.09________m

(d) The angle at t = 2.0 s.

____0.480___________rad

.