Optics or Classical Mechanics

Mayank and Akul had their birthdays last Friday. They got 2 spotlights, a detector and a spring in their birthday presents. They thought of doing something crazy.

Akul welded the spotlights at the ends of the spring and Mayank took the detector and ran around in a circle of very large radius with its centre coinciding with the centre of spring as shown.

At t = 0 t=0 , spring was in its r e l a x e d relaxed s t a t e state , Mayank was at θ = 0 \theta =0 and Akul gave the spring an impulse, such that it starts oscillating. Mayank changed his position with time in such a way that he A L W A Y S ALWAYS detected a f i r s t first order m a x i m a maxima .

It is known that:-

1) Mayank moved from θ ϵ [ 0 , π 4 ] \theta \epsilon \left[ 0,\frac { \pi }{ 4 } \right]

2) The mass of spotlights is 1 k g 1kg and spring constant is 1 N / m 1N/m .

3) Always means at each and every instant of time.

4) Akul gives impulse such that the spring expands at t=0 i.e. outward impulse.

Find Mayank's angular position as a function of time. Give your answer(in degrees) as his position at t = 1 t=1 second.

Wanna have more fun with Mayank and Akul. This question is a part of the set Mayank and Akul

12.56 33.12 44.79 30 23.11 41.21

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1 solution

Sanchit Aggarwal
Oct 4, 2015

Exact answer:- 44.29 degrees. Please change it.

No it is 44.79 Guide me, why you think so

Mayank Singh - 5 years, 8 months ago

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question is wrong as it has not been mentioned if the sources are coherent

Avinash S - 5 years, 5 months ago

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I would like to state here that 90% of the questions in IIT(or any institution's question paper) don't mention this thing. What would you do there? Would you solve or mark the answer as None of These?(I didn't even provide)

What follows is obvious

Mayank Singh - 5 years, 5 months ago

can u please post the solution to this question or even a hint

Avinash S - 5 years, 5 months ago

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@Avinash S see that the springs perform SHM (say with an amplitude A), now you can write the length of the spring as a function of time(initial length must be lambda)

now apply the condition of first order maxima for Mayank's movements.

Since the range of Theta is limited to 45 degrees, you can use it to find the amplitude A

Mayank Singh - 5 years, 5 months ago

because that's what Wolfram calculated.

Sanchit Aggarwal - 5 years, 8 months ago

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Your equations might have went wrong... Please provide a detailed solution

Mayank Singh - 5 years, 8 months ago

My answer also came 44.285 to be exact!

Aniket Sanghi - 4 years, 8 months ago

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