The length of a simple pendulum is 0.70 m and the mass of the parade Cthe \"bob\

Question

The length of a simple pendulum is 0.70 m and the mass of the parade Cthe "bob") at tne end of the cable is 0.37 kg. The pendulum is pulled away from its equilibrium position by an angle of 6.8 degree and released from rest. Assume that friction can be neglectd and that the resulting oscillatory motion is simple harmonic motion.(a) What is the angular frequency of the motion? (b) Using the position of the bob at Its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth, (c) What Is tne bob's speed as it passes truouofi the lowest point ol the swing?

Explanation / Answer

T = 2*pi*sqrt(L/g) = 2*3.14*sqrt(0.7/9.8)= 1.678 s

W = 2*pi/T = 3.74 rad/s

b) TE = m g h = m*g*L*(1-cos6.8) = 0.37*9.8*0.7*(1-cos6.8) = 0.0178 J

c) TE = 0.5*M*V^2

V = 0.311 m/s

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