In: Physics
A simple pendulum with mass m = 1.3 kg and length L = 2.62 m hangs from the ceiling. It is pulled back to an small angle of ? = 11.6° from the vertical and released at t = 0.
What is the period of oscillation?
What is the magnitude of the force on the pendulum bob perpendicular to the string at t=0?
What is the maximum speed of the pendulum?
What is the angular displacement at t = 3.67 s? (give the answer as a negative angle if the angle is to the left of the vertical)
What is the magnitude of the tangential acceleration as the pendulum passes through the equilibrium position?
What is the magnitude of the radial acceleration as the pendulum passes through the equilibrium position?
(a) period of oscillation T = 2*pi* squareroot(L/g);
T = 2*pi*squareroot(2.62/9.8);
T = 3.248 sec.
(b) magnitude of the force on the pendulum bob perpendicular to the string at t=0;
tangential force Ft = mg*sin(theta);
Ft = 1.3*9.8*sin(11.6);
Ft = 2.56 N.
(c) maximum speed of the pendulum;
mgh = 1/2(mv2); (at the bottom potential energy tranformed into Kinetic energy)
where h = L - L*cos(theta) = L(1 - cos(theta)) = 2.62*(1 - cos(11.6));
h = 0.0535 m.
v = squareroot(2*g*h);
v = squareroot(2*9.8*0.535);
v = 1.024 m/s (max speed).
(d) angular displacement at t = 3.67 s;
angular frequency w = squareroot(g/L) = 1.934 /sec;
thetamax = 11.6 degrees;
wt = 1.934*3.67 = 7.1 radians = 7.1*180/pi = 406.678 degrees;
theta = thetamax(cos(wt));
theta = 11.6*cos(406.678)
theta = 7.958 degrees. (angular displacement at t = 3.67 s)
(e) What is the magnitude of the tangential acceleration at the bottom;
at the bottom tangential foce will be zero so tangential acceleration will also zero.
(f) magnitude of the radial acceleration as the pendulum passes through the equilibrium position:
radial acceleration = v2 / L;
= (1.024)2 / 2.62;
= 0.4 m/s2.
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Which of the following would change the frequency of oscillation of this simple pendulum?
increasing the mass
decreasing the initial angular displacement
increasing the length (correct)
hanging the pendulum in an elevator accelerating downward (correct)