A pendulum of length L_1 has a period T_1 = 0.910 s The length of the pendulum i

Question

A pendulum of length L_1 has a period T_1 = 0.910 s The length of the pendulum is adjusted to a new value L_2 such that T_2 = 1.00 s. What is the ratio L_2/L_1? A grandfather clock is constructed so that it has a simple pendulum that swings from one side to the other, a distance of 21.0 mm, in 1.00 s. What is the maximum speed of the pendulum bob? Use two different methods. First, assume SHM and use the relationship between amplitude and maximum speed. Because of dissipative forces, the amplitude of an oscillator decreases 8.7% in 9 cycles. How many cycles does it take for the energy to decrease 8.7%?

Explanation / Answer

4) For a pendulum omega (angular frequency) = 2*pi/T where T is the period Here T = 2.0s (time for one complete oscillation) so omega = pi = 3.14 rad/s. From the eqns for SHM you can derive the function for vmax.... vmax = omega*A where A is the amplitude (In this problem A = 0.021/2m)

So vmax = 3.14*0.021/2 = 0.0361 m/s = 3.30cm/s

B) Here (K +U)bottom = (K + U) top Note U bot = 0 and K top = 0

So K bot =1/2*m*vmax^2 = m*g*y or vmax = sqrt(2*g*y) All we need is y. The length of a pendulum with a period of 2.0s is T^2*g/(4*pi^2) = 0.993m

So the angle between the bottom and the top can be found using s = R*theta theta = s/R
= 0.021/2/.993 = 0.01057rad = 0.6056deg


Therefore the height of the bob is 0.993*(1-cos(.6635)) = 5.547x10^-5 m

So vmax = sqrt(2*5.547x10^-5*9.8) = 3.30cm/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.