What is the period of a pendulum formed by placing a horizontal axis (a) through the end of a meterstick ( 100 -cm mark)? (b) through the 75 -cm mark? (c) through the \(60-\mathrm{cm}\) mark?

Short Answer

Expert verified
Answer: The approximate periods of the pendulum for the given cases are: Case a: 2.006 s, Case b: 1.736 s, and Case c: 1.554 s.

Step by step solution

01

(Case a: Horizontal Axis at 100-cm mark)

For this case, the horizontal axis is placed at the 100-cm mark. So the length of the pendulum, \(L\), will be 1 meter (100 cm). First, we convert the length into meters: $$L = 100\, \mathrm{cm} \times \frac{1\, \mathrm{m}}{100\, \mathrm{cm}} = 1\, \mathrm{m}$$ Now we can calculate the period, \(T\), using the formula: $$ T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{1\, \mathrm{m}}{9.81\, \mathrm{m/s^2}}}$$ After plugging the values, the period for case a is approximately: $$ T \approx 2.006\, \mathrm{s}$$
02

(Case b: Horizontal Axis at 75-cm mark)

For this case, the horizontal axis is located at the 75-cm mark. So the length of the pendulum, \(L\), will be 0.75 meters. First, convert the length into meters: $$L = 75\, \mathrm{cm} \times \frac{1\, \mathrm{m}}{100\, \mathrm{cm}} = 0.75\, \mathrm{m}$$ Now we can calculate the period, \(T\), using the formula: $$ T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{0.75\, \mathrm{m}}{9.81\, \mathrm{m/s^2}}}$$ After plugging the values, the period for case b is approximately: $$ T \approx 1.736\, \mathrm{s}$$
03

(Case c: Horizontal Axis at 60-cm mark)

For this case, the horizontal axis is located at the 60-cm mark. So the length of the pendulum, \(L\), will be 0.6 meters. First, convert the length into meters: $$L = 60\, \mathrm{cm} \times \frac{1\, \mathrm{m}}{100\, \mathrm{cm}} = 0.6\, \mathrm{m}$$ Now we can calculate the period, \(T\), using the formula: $$ T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{0.6\, \mathrm{m}}{9.81\, \mathrm{m/s^2}}}$$ After plugging the values, the period for case c is approximately: $$ T \approx 1.554\, \mathrm{s}$$

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