A simple pendulum is 0.26 mlong. At t = 0 it is released from rest startingat an

Question

A simple pendulum is 0.26 mlong. At t = 0 it is released from rest startingat an angle of 11°. Ignore friction. (Use g =9.8 m/s2.)
(a) What will be the angular position of the pendulumat t = 0.65 s?
°

(b) What will be the angular position of the pendulumat t = 1.95 s?
°

(c) What will be the angular position of the pendulumat t = 5.00 s?
°
(a) What will be the angular position of the pendulumat t = 0.65 s?
°

(b) What will be the angular position of the pendulumat t = 1.95 s?
°

(c) What will be the angular position of the pendulumat t = 5.00 s?
°

Explanation / Answer

Although there are three questions to this problem, all we reallyneed to do is find a function for in terms of t and we canjust plug in the necessary three values for t to get our threeanswers. The equation for the angular position of a simple pendulum startingfrom rest at max is given via simple harmonic motion: (t) = 0*cos(2t / T) Where 0 is the maximum angle that the pendulum canachieve, t is the time passed, and T is the period ofoscillation. We can find T using another equation for simplependulums: T = 2*(L / g) The problem statement gives us the necessary info to solve forT: T = 2*(0.26 / 9.8) T = 1.023 s Now we can formulate our function for (remember that0 = 11 deg): (t) = 11*cos(2t / 1.023)(specific to this problem only) Remember to set your calculator to degree mode. I'll do the first one: t = 0.65 s: (t) = 11*cos(2*0.65 / 1.023) (t) = 11 degrees (rounded to two sigfigs). This is because the pendulum has made just about oneswing over to its peak on the other side. The other two are very similar.

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