
A 2.0 kg particle undergoes SHM according to $x=1.5\sin \left( \dfrac{\pi t}{4}+\dfrac{\pi }{6} \right)$ ( in SI units). What is the total mechanical energy of the particle?
Answer
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Hint:In this question, we have to find the mechanical energy of the particle. To find the energy, first we compare the given equation with the standard form of SHM and then by putting the values in the formula of mechanical energy and by solving equations, we are able to find the value and choose the correct option.
Formula Used:
Mechanical energy can be found using formula = $\dfrac{1}{2}m{{w}^{2}}{{A}^{2}}$
Where w is the angular velocity of SHM, A is the amplitude of SHM and m is the mass of the particle.
Complete step by step solution:
Given equation is $x=1.5\sin \left( \dfrac{\pi t}{4}+\dfrac{\pi }{6} \right)$
And the mass of the particle = 2.0 kg
Comparing the given equation with $x(t)=A\sin (wt+{{\phi }_{0}})$
We get $A=1.5m$, $w=\dfrac{\pi }{4}rad/s$, ${{\phi }_{0}}=\dfrac{\pi }{6}$
As we know the time period is
$T=\dfrac{2\pi }{w}=8s$
Total energy = $\dfrac{1}{2}m{{v}^{2}}_{\max }$
As, $v=wA$
Therefore,
E = $\dfrac{1}{2}m{{w}^{2}}{{A}^{2}}$
Then
$E = \dfrac{1}{2}\times 2.0\times {{\left( \dfrac{\pi }{4} \right)}^{2}}\times {{(1.5)}^{2}}$
Solving further the above equation, we get
$E = 1.39\,J$
Hence, total mechanical energy of the particle = 1.40 J
Note: Simple harmonic motion involves an interconnection between different types of energy. For example- when a horizontal spring oscillates back and forth, then both elastic potential and kinetic energy comes into play. When a pendulum swings, then gravitational potential energy and kinetic energy comes into play. And when a vertical spring is oscillating, both gravitational and elastic potential energy are involved with kinetic energy in the middle comes into play.
Formula Used:
Mechanical energy can be found using formula = $\dfrac{1}{2}m{{w}^{2}}{{A}^{2}}$
Where w is the angular velocity of SHM, A is the amplitude of SHM and m is the mass of the particle.
Complete step by step solution:
Given equation is $x=1.5\sin \left( \dfrac{\pi t}{4}+\dfrac{\pi }{6} \right)$
And the mass of the particle = 2.0 kg
Comparing the given equation with $x(t)=A\sin (wt+{{\phi }_{0}})$
We get $A=1.5m$, $w=\dfrac{\pi }{4}rad/s$, ${{\phi }_{0}}=\dfrac{\pi }{6}$
As we know the time period is
$T=\dfrac{2\pi }{w}=8s$
Total energy = $\dfrac{1}{2}m{{v}^{2}}_{\max }$
As, $v=wA$
Therefore,
E = $\dfrac{1}{2}m{{w}^{2}}{{A}^{2}}$
Then
$E = \dfrac{1}{2}\times 2.0\times {{\left( \dfrac{\pi }{4} \right)}^{2}}\times {{(1.5)}^{2}}$
Solving further the above equation, we get
$E = 1.39\,J$
Hence, total mechanical energy of the particle = 1.40 J
Note: Simple harmonic motion involves an interconnection between different types of energy. For example- when a horizontal spring oscillates back and forth, then both elastic potential and kinetic energy comes into play. When a pendulum swings, then gravitational potential energy and kinetic energy comes into play. And when a vertical spring is oscillating, both gravitational and elastic potential energy are involved with kinetic energy in the middle comes into play.
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