
At a time when the displacement is $\dfrac{1}{2}$ the amplitude. What fraction of energy is K.E. and what fraction is potential energy in S.H.M.
Answer
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Hint:First of all, we have to find the total amount of energy associated when the displacement is half the amplitude. Then we have to find the amount of potential energy involved in this case. Then by using potential energy we will find the kinetic energy.
Complete step by step answer:
Let the total energy associated with a S.H.M. be $U$. The total energy is thus given as,
$U = \dfrac{1}{2}km_m^2$.
The variables are defined as, $k = $ force constant of a spring and ${x_m} = $ maximum amplitude.
Now, let us assume that $x$ is the displacement here. According to the question we have to find the Kinetic energy and Potential Energy when the displacement is half of the amplitude.
$x = \dfrac{1}{2}{x_m}$.
So, the potential energy $P$ can be expressed as,
$P = \dfrac{1}{2}k{x^2} \\
\Rightarrow P= \dfrac{1}{2}k{\left( {\dfrac{1}{2}{x_m}} \right)^2}$
Thus, now we find the value of Potential Energy $P = \dfrac{1}{8}kx_m^2$.
The Kinetic Energy $K$ can be calculated as the energy left from the subtraction of Potential Energy from Total Energy,
$K = U - P$
$\Rightarrow K = \dfrac{1}{2}km_m^2 - \dfrac{1}{8}kx_m^2 \\
\Rightarrow K= \dfrac{3}{8}kx_m^2$
So, we find kinetic energy as,
$K = \dfrac{3}{8}kx_m^2$
Now, the fraction of Kinetic Energy with respect to Total energy is,
$\dfrac{K}{U} = \dfrac{{\dfrac{3}{8}kx_m^2}}{{\dfrac{1}{2}kx_m^2}} \\
\Rightarrow \dfrac{K}{U} = \dfrac{3}{4}$
Now, the fraction of Potential energy with respect to Total Energy is,
$\dfrac{P}{U} = \dfrac{{\dfrac{1}{8}kx_m^2}}{{\dfrac{1}{2}kx_m^2}} \\
\therefore \dfrac{P}{U}= \dfrac{1}{4}$
Therefore, we conclude that $\dfrac{3}{4}$ of energy is Kinetic energy and $\dfrac{1}{4}$ of energy is Potential energy.
Note: It must be noted that potential energy of a substance is the stored energy that the body possesses. Kinetic energy refers to the energy associated with a particle when it is in a moving state. Energy is always conserved, it changes its forms and can never be created or destroyed.
Complete step by step answer:
Let the total energy associated with a S.H.M. be $U$. The total energy is thus given as,
$U = \dfrac{1}{2}km_m^2$.
The variables are defined as, $k = $ force constant of a spring and ${x_m} = $ maximum amplitude.
Now, let us assume that $x$ is the displacement here. According to the question we have to find the Kinetic energy and Potential Energy when the displacement is half of the amplitude.
$x = \dfrac{1}{2}{x_m}$.
So, the potential energy $P$ can be expressed as,
$P = \dfrac{1}{2}k{x^2} \\
\Rightarrow P= \dfrac{1}{2}k{\left( {\dfrac{1}{2}{x_m}} \right)^2}$
Thus, now we find the value of Potential Energy $P = \dfrac{1}{8}kx_m^2$.
The Kinetic Energy $K$ can be calculated as the energy left from the subtraction of Potential Energy from Total Energy,
$K = U - P$
$\Rightarrow K = \dfrac{1}{2}km_m^2 - \dfrac{1}{8}kx_m^2 \\
\Rightarrow K= \dfrac{3}{8}kx_m^2$
So, we find kinetic energy as,
$K = \dfrac{3}{8}kx_m^2$
Now, the fraction of Kinetic Energy with respect to Total energy is,
$\dfrac{K}{U} = \dfrac{{\dfrac{3}{8}kx_m^2}}{{\dfrac{1}{2}kx_m^2}} \\
\Rightarrow \dfrac{K}{U} = \dfrac{3}{4}$
Now, the fraction of Potential energy with respect to Total Energy is,
$\dfrac{P}{U} = \dfrac{{\dfrac{1}{8}kx_m^2}}{{\dfrac{1}{2}kx_m^2}} \\
\therefore \dfrac{P}{U}= \dfrac{1}{4}$
Therefore, we conclude that $\dfrac{3}{4}$ of energy is Kinetic energy and $\dfrac{1}{4}$ of energy is Potential energy.
Note: It must be noted that potential energy of a substance is the stored energy that the body possesses. Kinetic energy refers to the energy associated with a particle when it is in a moving state. Energy is always conserved, it changes its forms and can never be created or destroyed.
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