When a kid moves up and down on a swing with very little effort, the swing is be

Question

When a kid moves up and down on a swing with very little effort, the swing is below its natural frequency, above its natural frequency, or at its natural frequency? And a person sits on a swing. When the person sits still, the swing oscillates back and forth at its natural frequency. If, instead, two people sit on the swing, the new natural frequency of the swing is greater than it was with one person, the same as it was with one person, or less than it was with one person?

Please explain the answer so I can make sure I fully understand it! Thank you (:

Explanation / Answer

(a) The swing oscillate with it's natural frequency.               f = (1 / 2) ( k / m ) (b) When person sits in the swing: Frequency is inversely proportional to the square root of mass. So, the frequency decreases due to increase in mass. (c) Two people sits in the string. Frequency is inversely proportional to the square root of mass. So, the frequency decreases due to increase in mass. It is less than it was when compared to one person. Frequency is inversely proportional to the square root of mass. So, the frequency decreases due to increase in mass. It is less than it was when compared to one person.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---