Question

A spring with a weight attached to its end begins to oscillate at a maximum amplitude of 30 cm. Due to the effect of air, the amplitude will decrease in time while the period remains constant. One way to study this is through the time taken for the amplitude to decrease by half, known as the decay half-time. If this system has a decay half-time of 20 seconds, calculate the amplitude of the oscillation one minute after it starts to oscillate.

a) 30 cm
b) 15 cm
c) 7.5 cm
d) 3.75 cm

Answer 1

The amplitude of the oscillation one minute after starting will be 3.75 cm, after going through three half-lives of decay, each reducing the amplitude by half from the starting amplitude of 30 cm.

Step-by-step explanation:

The question revolves around the concept of damped harmonic motion, where the amplitude of oscillation decays over time due to resistive forces like air friction. Given that the spring has a decay half-time of 20 seconds, we can determine the amplitude of oscillation after one minute (which is three half-times of 20 seconds each).

Starting with an initial amplitude of 30 cm, after each half-time, the amplitude will be halved. So after:

• 20 seconds, the amplitude would be halved to 15 cm,
• 40 seconds (two half-times), it would be halved again to 7.5 cm,
• 60 seconds (three half-times), it would be halved once more to 3.75 cm.

Therefore, the amplitude of the oscillation one minute after it starts to oscillate would be 3.75 cm, which corresponds to choice (d) 3.75 cm.