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TOPIC 7: Oscillations and Resonance
Although not officially in the syllabus, the following equations should be memorised. The alternative, ie. to derive on the spot, may be difficult for the average student.
For a spring-mass system, the natural period of oscillation is given by:
T = 2p Ö(m/k) = 2p Ö(e/g).
where m is the mass of the object attached to the spring; k the spring constant; e the extension of the spring; and g the acceleration due to gravity.
Why is m/k = e/g? That's because mg = ke.
For a pendulum system, the natural period of oscillation is given by:
T = 2p Ö(l/g).
where l is the length of the string and g the acceleration of free fall.
For a spring-mass system, if the the spring is replaced by 2 springs of the same kind in series, and the mass is doubled, what happens to the period of oscillation?
Answer: For 2 springs in series, the spring constant is halved. Since m is doubled and k is halved, m/k will quadruple; and Ö(m/k) will double. Therefore, the period T will double.
If I relocate both a pendulum and a spring-mass system to the moon, where the acceleration due to gravity is 1/6 that on Earth, what happens to the period in each case?
Answer: the period of the pendulum will be Ö6 times that on Earth, while that of the spring-mass system will remain unchanged.
Reason: since g becomes 1/6 times its original value, the ratio l/g becomes 6 times its original value. Therefore, Ö(l/g) becomes Ö6 times its original value.
But what about the spring-mass system? Well, in this case, the mass m is constant (mass does not change); the spring constant k is constant. Hence period T = 2p Ö(m/k) is also constant.
But for a spring-mass system, isn't period T also = 2p Ö(e/g) ? So shouldn't the period also be affected by g as well?
Answer: on the moon, g is 1/6 times, but the extension of the spring also becomes 1/6 times since the mass now weighs only 1/6 of its weight on Earth. So the 1/6 in the numerator cancels out with the 1/6 in the denominator, and the period remains unchanged.
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