Question

Gillian buys a pendulum clock at a discount store and discovers when she gets it home that it loses 6.00 minutes each day.

a) Should she lengthen or shorten the pendulum in order for it to keep accurate time?
b) If the pendulum has a period of 2.00 s, by how much must the length be changed so that the clock keeps accurate time?

Answer 1

Answer:

a) The pendulum must be shortened to keep accurate time

b) It should be shortened by 0.0082 m or 8.2 mm

Explanation:

Simple Pendulum

A simple pendulum is a system with a point mass that is suspended from a weightless string to a fixed point. It describes a harmonic motion because the oscillations repeat regularly, and kinetic energy is transformed into potential energy, and vice versa.

The equation for the period of a simple pendulum is

$\displaystyle T = 2\pi \sqrt{\frac{L}{ g}}$

where L is the length of the sting and g is the acceleration of gravity .

Note that if we increase the length, the period will also increase, and the oscillations are slower, i.e. take longer to complete

a) Gillian's pendulum is running slow because it loses 6 minutes each day. As shown above, the longer the string, the slower the oscillations, thus he needs to shorten the string to make it move faster and keep up in time .

b) We know the period is 2 seconds. That will give us the actual length of the pendulum, solving the above equation for L .

$\displaystyle L = \frac{gT^2}{4\pi^2}$

$\displaystyle L = \frac{(9.8)2^2}{4\pi^2}$

$\displaystyle L = 0.9929\ m$

The new period should be less than the original. We know that actually, the pendulum's mechanisms make a period of 2 seconds in a measured time of

3600*24+6*60=86760 seconds each day .

This should be shortened to the correct time of 3600*24=86400 seconds per day, so the new period should be

$T'=2\ sec*(86400/86760)=1.9917\ sec$

Which yields to a new length  of

$\displaystyle L' = \frac{gT'^2}{4\pi^2}$

$\displaystyle L' = \frac{(9.8)(1.9917)^2}{4\pi^2}$

$L'=0.9847\ m$

Difference of lengths = 0.9929 - 0.9847=0.0082 m

$\boxed{\text{It should be shortened by 0.0082 m or 8.2 mm}}$

Answer 2

When the pendulum loses time, the length of the pendulum can be increased to account for the lost time.

The length must be changed by 358.5 m so that the pendulum will keep accurate time.

Period of oscillation

The period of oscillation of a simple pendulum is calculated as follows;

$T = 2\pi \sqrt{\frac{l}{g} }$

The period of oscillation is directly proportional to the length of the pendulum.

When the pendulum loses time, the length of the pendulum can be increased to account for the lost time.

Length when period is 2 seconds

$\frac{T^2}{4\pi ^2} = \frac{l}{g} \\\\l = \frac{T^2 g}{4\pi^2}\\\\l =\frac{2^2 \times 9.8}{4\pi^2} \\\\l = 0.992 \ m$

Since the pendulum losses 6 minutes

The actual time = 6 mins + 2s = (6 x 60) + 2 = 362 s

Length for the actual time of the pendulum

$\frac{T^2}{4\pi ^2} = \frac{l}{g} \\\\l = \frac{T^2 g}{4\pi^2}\\\\l =\frac{(362)^2 \times 9.8}{4\pi^2} \\\\l = 32,529.96 \ m$

Length for the lost time (6mins)

$\frac{T^2}{4\pi ^2} = \frac{l}{g} \\\\l = \frac{T^2 g}{4\pi^2}\\\\l =\frac{360^2 \times 9.8}{4\pi^2} \\\\l = 32,171.5 \ m$

difference in length = 32,529.96 - 32,171.5 = 358.5 m

Learn more about period of pendulum here: https://brainly.com/question/26449711