Answer to Question #165647 in Physics for Vic

Question

We will look at the motion of a simple pendulum consisting of a bob with mass 0.450

kg tied to the ceiling by a string with length 1.20 m.

1. What is the period and frequency of the motion of this pendulum?

2. Suppose that the length of the string is halved.

a. What is the period of the pendulum given this shorter length?

b. By how much did the period of the pendulum change from its original

length?

c. If the length of the pendulum was instead doubled, what would be the

new period of the pendulum?

3. The original pendulum is brought and made to swing on the surface of the

moon. Acceleration due to gravity on the moon is equal to g = 1.62 m/s^2.

a. What is the period and frequency of the pendulum on the moon?

b. By how much did the period of the pendulum change compared to when it

was on Earth?

4. Suppose that we are to construct a spring system with the same period of

oscillation as the simple pendulum in (1). What must be the spring coefficient of

the spring if we are to use the same 0.450-kg mass as load?

Answer 1

1)

$$T=2\pi\sqrt{\dfrac{L}{g}}=2\pi\sqrt{\dfrac{1.2\ m}{9.8\ \dfrac{m}{s^2}}}=2.2\ s,$$ $$f=\dfrac{1}{T}=\dfrac{1}{2.2\ s}=0.45\ Hz.$$

2) a)

$$T_{new}=2\pi\sqrt{\dfrac{L}{2g}}=\dfrac{1}{\sqrt{2}}\cdot2\pi\sqrt{\dfrac{L}{g}}=\dfrac{T}{\sqrt{2}},$$ $$T_{new}=\dfrac{2.2\ s}{\sqrt{2}}=1.55\ s.$$

b) The period of the pendulum with halved length of string becomes $1/\sqrt{2}$ times shorter than the period of the pendulum with original length.

c)

$$T_{new}=2\pi\sqrt{\dfrac{2L}{g}}=\sqrt{2}\cdot2\pi\sqrt{\dfrac{L}{g}}=\sqrt{2}T,$$ $$T_{new}=\sqrt{2}\cdot2.2\ s=3.11\ s.$$

The period of the pendulum with doubled length of string becomes $\sqrt{2}$ times longer than the period of the pendulum with original length.

3) a)

$$T=2\pi\sqrt{\dfrac{L}{g}}=2\pi\sqrt{\dfrac{1.2\ m}{1.62\ \dfrac{m}{s^2}}}=5.4\ s,$$ $$f=\dfrac{1}{T}=\dfrac{1}{5.4\ s}=0.185\ Hz.$$

b)

$$T_{new}=2\pi\sqrt{\dfrac{L}{\dfrac{1}{6}g}}=\sqrt{6}\cdot2\pi\sqrt{\dfrac{L}{g}}=\sqrt{6}T=2.45T.$$

The period of the pendulum on Moon becomes 2.45 times longer than the period of the pendulum on Earth.

4)

$$T=2\pi\sqrt{\dfrac{m}{k}},$$ $$k=\dfrac{4\pi^2m}{T^2}=\dfrac{4\pi^2\cdot0.45\ kg}{(2.2\ s)^2}=3.67\ \dfrac{N}{m}.$$