| 1) (a) What must be the length of a pendulum to produce a period of 1.0 s? |
| (b) How would you modify a pendulum to produce a period of 1.0 s on the surface of the moon, where |
| gm = 1/6 x ge? |
| (a) T = 1.0 s g = 9.80 m/s2 |
| T = 2π(l/g)1/2 |
| l = T2 x g/4π2 = (1.0 s)2 x 9.80 m/s2/(4 x π2) = 0.25 m |
| (b) T = 1.0 s gm = 1/6 x 9.80 m/s2 |
| l = T2 x g/4π2 = (1.0 s)2 x 1/6 x 9.80 m/s2/(4 x π2) = 0.041 m |
| Because g was not specified, g could have been expressed as 980 cm/s2 or 32 ft/s2. |
| 2) (a) Determine the period and frequency of the pendulum in the animation. |
| (b) When is the pendulum moving the fastest and what is its acceleration at that point? |
| (a) Using the control keys (Play, Pause, etc.), the period, T, is determined to be 10.4 s (the time needed for |
| one complete vibration). |
| The frequency, f, equals: F = T-1 = (10.4 s)-1 = 0.0962 s. |
| (b) The pendulum is moving the fastest as it moves through the equilibrium position. At the equilibrium |
| position, the restoring force, Fr = 0, making the acceleration equal to zero. |
| 3) Why doesn't the mass of the pendulum bob affect the period of a pendulum? |
| The greater the mass, the greater the weight, which makes the restoring force larger, Fr = Fwsinθ. |
| However, a larger mass needs a larger force to achieve the same acceleration. This is analogous to a |
| heavier and a lighter object in free fall. |
| 4) Why doesn't the amplitude affect the period of a pendulum for small angles? |
| When the amplitude is increased, the restoring force is also increased. The larger restoring force results |
| in a greater acceleration but the distance is also greater. |
| 5) Using the conservation of energy, show that vmax = (2gh)1/2. |
| ΔE = 0 |
| (KE + PEg)T = (KE + PEg)B |
| 0 + mgh = 1/2mv2 + 0 |
| v = (2gh)1/2 |