To confirm my answer wheter is right or wrong

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SUMMARY

The maximum kinetic energy (Kmax) of a mass m attached to a vertical spring with spring constant k, undergoing simple harmonic motion with amplitude a, is definitively expressed as Kmax = π² * m * a² * k. This formula derives from the relationship between kinetic energy and the angular frequency of the motion. The correct expression for maximum kinetic energy incorporates the amplitude and spring constant, confirming the solution provided in the discussion.

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Sanosuke Sagara
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A mass m at the end of a vertical spring constant k vibrates vertically with simple harmonic motion of amplitude a.Find the expression for the maximum kinetic energy of the mass in terms of m,k and a.


My solution :
 

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i think your maximum kinetic energy should be 1/2*m*v^2 = 1/2*m*w^4*x^2 in stead of 1/2*m*w^2*x^2
 
Last edited:


Your solution is correct. The maximum kinetic energy of the mass can be expressed as:

Kmax = 1/2 * m * (2π * a * √(k/m))^2

or in simplified form as:

Kmax = π^2 * m * a^2 * k

Great job on solving this problem!
 

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