SHM!

Classical Mechanics Level 3

A block is performing SHM of amplitude A A in vertical direction. When block is at y y (measured from mean position), it detaches from spring, so that the spring contracts and does not affect the motion of the block. Find y ∗ { y }^{ * } such that the bock attains maximum height from the mean position.

Given - A ω 2 > g A{\omega}^{2}>g , where ω \omega is the angular frequency and g g is acceleration due to gravity.

More questions??

g 2 ω 2 \frac { { g }^{ 2 } }{ { \omega }^{ 2 } } g ω 2 \frac { g }{ { \omega }^{ 2 } } g 2 ω \frac { { g }^{ 2 } }{ { \omega } } g ω \frac { g }{ { \omega } }

Anandhu Raj by Anandhu Raj , 23, India

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2 solutions

Neelesh Vij
Mar 28, 2016

An improper method (but reasonable) is that to check dimensions, of the options given. Thus only one option has the correct dimensions thus the answer is g w 2 \boxed {\dfrac{g}{w^2}}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Aw!! I should have checked that while making options! Silly me! :P :)

Anandhu Raj - 5 years, 2 months ago

Did the same! :) Otherwise its a one liner

mw^2 = mg/y

Prakhar Bindal - 5 years, 2 months ago
Md Zuhair
Nov 16, 2017

Why is it level 4!! HAHA!!

Its

m g = m ω 2 y mg=m \omega^{2} y

⟹ y = g ω 2 \boxed{\boxed{\boxed{\boxed{\boxed{\implies y=\dfrac{g}{\omega^2}}}}}}

Else easiest way is matching the dimensions of the options!

0 Helpful 0 Interesting 0 Brilliant 0 Confused

It's level 4 because nobody has tried such an easy question

Kumar Krish - 2 years, 4 months ago

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