A student performs an experiment that involves the motion of a pendulum. The student attaches one end of a string to an object of mass M and secures the other end of the string so that the object is at rest as it hangs from the string. When the student raises the object to a height above its lowest point and releases it from rest, the object undergoes simple harmonic motion. As the student collects data about the time it takes for the pendulum to undergo one oscillation, the student observes that the time for one swing significantly changes after each oscillation. The student wants to conduct the experiment a second time. Which two of the following procedures should the student consider when conducting the second experiment?
a) Make sure that the length of the string is not too long.
b) Make sure that the mass of the pendulum is not too large.
c) Make sure that the difference in height between the pendulum's release position and rest position is not too large.
d) Make sure that the experiment is conducted in an environment that has minimal wind resistance.

In this simple pendulum experiment the student observes a significant change in time between each period. This occurs since an approximation used is that the sine of the angle is small, so

sin θ = θ

with this approach the equation will be surveyed

d² θ / dt² = - g / L sin θ

It is reduced to

d² θ / dt² = - g / L θ

in which the time for each oscillation is constant, for this approximation the angle must be less than 10º so that the difference between the sine and the angles is less than 1%

The angle is related to the height of the pendulum

sin θ = h / L

h = L sin θ.

Therefore the student must be careful that the height is small.

When reviewing the answers the correct one is cη

pristina pristina · Answer 2 · 2022-02-25T15:58:57+01:00

Considering the approximation of simple harmonic motion, the correct option is:

(c) Make sure that the difference in height between the pendulum's release position and rest position is not too large.

Simple Harmonic Motion

According to Newton's second law in case of rotational motion, we have;

[tex]\tau = I \alpha[/tex]

Applying this, in the case of a simple pendulum, we get;

[tex]-mg\,sin\,\theta =mL^2 \,\frac{d^2 \theta}{dt^2}[/tex]

On, rearranging the above equation, we get;

[tex]mL^2 \,\frac{d^2 \theta}{dt^2} + mg\,sin\,\theta=0\\\\\implies \frac{d^2 \theta}{dt^2} +\frac{g}{L} sin \,\theta=0[/tex]

Now, if angular displacement is very small, i.e.; the bob of the pendulum is only raised slightly.

Then, [tex]sin\, \theta \approx \theta[/tex]

[tex]\implies \frac{d^2 \theta}{dt^2} +\frac{g}{L} \,\theta=0[/tex]

This is now in the form of the equation of a simple harmonic motion.

[tex]\frac{d^2 \theta}{dt^2} +\omega^2 \,\theta=0[/tex]

Comparing both these equations, we can say that;

[tex]\omega = \sqrt{\frac{g}{L}}[/tex]

[tex]T=2\pi\sqrt{\frac{L}{g}}[/tex]

This relation for the time period can only be obtained if the angular displacement is very less.

So, the correct option is;

Option (c): Make sure that the difference in height between the pendulum's release position and rest position is not too large.

Learn more about simple harmonic motion here:

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