Question

Let’s think back to the lab for simple harmonic motion. Consider the setup for the simple...

Let’s think back to the lab for simple harmonic motion. Consider the setup for the simple pendulum. The length of the pendulum is 0.60 m and the bob has inertia 0.50 kg (assume mass of the string is negligible, and small angular displacements). You conduct two experiments (A and B) to investigate the physics of simple harmonic motion. For experiment A, you pull the pendulum 5 ◦ , or π 36 radians. In experiment B, the angular displacement is 10◦ , or π 18 radians. Define clockwise rotation as positive.

(e) What are the maximum kinetic energies and potential energies for both experiments?

(f) In general, where do these occur for the motion of a pendulum?

(g)How do the total energies compare for the two experiments?

(h) For experiment A, write the equation of motion for angular displacement, θ(t) =??? (use either a sine or cosine and remember that clockwise is positive). Let t = 0 be the instant the bob passes though equilibrium moving in the counter-clockwise direction.

Homework Answers

Answer #1

= 5o

= 10o

m =0.5 kg, l= 0.6m

e) P.Emax = m*g*hmax

cos() = x/l => x = l*cos()

=> hmax = l-x =l - l*cos()

for case 1(5o ):

hmax = l-x =l - l*cos() = 0.6 - 0.6*cos(5o) = 2.283*10^-3m

P.Emax=0.5*9.81*2.283*10^-3=0.011198J

According conservation of energy, at P.Emax is converted into K.Emax at h=0

K.Emax = 0.011198J

for case2(10o):

hmax = l-x =l - l*cos() = 0.6 - 0.6*cos(10o) =9.115*10^-3m

P.Emax=0.5*9.81*9.115*10^-3=0.0447J

K.Emax =0.0447J

f) P.Emax at extreme ends where h= hmax and K.E=0 (v=0)

and K.Emax at equilibrium position(h=0) where P.E= 0 and KE=max

g) E1/E2 = 0.2505

h)

=  − g *sin θl = -9.81*0.0871/0.6 =-1.42499

θo(t=0) =0

θ(t) =1/2* (-1.42499 )*t^2

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