Figure 2 (left) shows the pendulum inside Marty’s clock, which consists of a mas

Q: Figure 2 (left) shows the pendulum inside Marty’s clock, which consists of a mas

T = 2*pi*(L/g) L = 0.996 T = 2*pi*sqrt(0.996/9.81) T = 2.002 s loss = 2.002-2 = 0.002 s (b) for T' = 2 2 = 2*pi*sqrt(L'/9.81) L' = 0.994 m h = L-L' = 0.996-0.994 = 0.002 m

ID: 1371253 • Letter: F

Question

Figure 2 (left) shows the pendulum inside Marty’s clock, which consists of a massless rod, and a copper cube with side length 0.03 m. A clock should make one cycle in exactly two seconds - a ”tick” and a ”tock” - but his clock runs slow. Use g = 9.81 m/s 2 . (a) How much time does his clock lose per day? (b) He figured he can fix this by adding a thin copper plate on top of the cube, as in Figure 2 (right). Find h, the thickness of the plate needed, assuming the plate has the same cross-sectional area as the copper cube.

0.9960 m 0.03 m

Explanation / Answer

T = 2*pi*(L/g)

L = 0.996

T = 2*pi*sqrt(0.996/9.81)

T = 2.002 s

loss = 2.002-2 = 0.002 s

(b)

for T' = 2

2 = 2*pi*sqrt(L'/9.81)

L' = 0.994 m

h = L-L' = 0.996-0.994 = 0.002 m

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Figure 2 (left) shows the pendulum inside Marty’s clock, which consists of a mas

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Figure 2 (left) shows the pendulum inside Marty’s clock, which consists of a mas

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