As we know that total energy of SHM is always constant
so here we know that
total energy = Kinetic energy + potential energy
so here we have
potential energy = 2/3E
so we will have kinetic energy as
[tex]KE = E - \frac{2}{3}E[/tex]
now we have
[tex]KE = \frac{1}{3}E[/tex]
now for the speed of the block
[tex]\frac{1}{2} mv^2 = \frac{1}{3}E[/tex]
by solving above equation we have
[tex]v = \sqrt{\frac{2E}{3m}}[/tex]
The object's velocity is √[(kA²)/(3m)]
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Hooke's Law states that the length of a spring is directly proportional to the force acting on the spring.
[tex]\boxed {F = k \times \Delta x}[/tex]
F = Force ( N )
k = Spring Constant ( N/m )
Δx = Extension ( m )
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The formula for finding Young's Modulus is as follows:
[tex]\boxed {E = \frac{F / A}{\Delta x / x_o}}[/tex]
E = Young's Modulus ( N/m² )
F = Force ( N )
A = Cross-Sectional Area ( m² )
Δx = Extension ( m )
x = Initial Length ( m )
Let us now tackle the problem !
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Given:
mass of the object = m
force constant = k
maximum displacement = A
total mechanical energy = E
potential energy = ²/₃E
Unknown:
velocity of the object = v = ?
Solution:
We will use conservation of energy as follows:
[tex]\texttt{Total Mechanical Energy} = E_p + E_k[/tex]
[tex]E = \frac{2}{3}E + \frac{1}{2}mv^2[/tex]
[tex]E - \frac{2}{3}E = \frac{1}{2}mv^2[/tex]
[tex]\frac{1}{3}E = \frac{1}{2}mv^2[/tex]
[tex]\frac{1}{3} (\frac{1}{2}kA^2) = \frac{1}{2}mv^2[/tex]
[tex]\frac{1}{3}kA^2 = mv^2[/tex]
[tex]v^2 = \frac{1}{3}kA^2 \div m[/tex]
[tex]\large {\boxed {v = \sqrt{ \frac{kA^2}{3m}}}}[/tex]
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Grade: College
Subject: Physics
Chapter: Elasticity
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Keywords: Elasticity , Diameter , Concrete , Column , Load , Compressed , Stretched , Modulus , Young
