Wave Mechanics Question 559

Question: A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency co. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time

Options:

A) at the mean position of the platform

B) for an amplitude of $ \frac{g}{{{\omega }^{2}}} $

C) for an amplitude of $ \frac{g}{{{\omega }^{2}}} $

D) at the highest position of the platform

Show Answer

Answer:

Correct Answer: B

Solution:

[b] For block A to move in S.H.M. $ mg-N=m{{\omega }^{2}}x $ where x is the distance from mean position For block to leave contact N = 0
$ \Rightarrow ,mg=m{{\omega }^{2}}x\Rightarrow x=\frac{g}{w^{2}} $



sathee Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on

Please select your preferred language
कृपया अपनी पसंदीदा भाषा चुनें