Question

$\huge\red{\tt Attention}$

A harmonic oscillator of force constant 4×10^6Nm^-1 and amplitude 0.01m has total energy 240J . What is maximum Kinetic Energy and minimum Potential Energy.​

Answer 1

$\huge\underline\blue{\sf Answer:}$

$\large\red{\boxed{\sf K_{max}=200J}}$

$\large\red{\boxed{\sf U_{min}=40J }}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Force constant (k) = $\sf{4×10^6Nm^{-1}}$

• Amplitude (A) = 0.01 m

• Total Energy (TE) = 240J

$\large\underline\pink{\sf To\:Find: }$

• Maximum Kinetic Energy ($\sf{K_{max}}$)=?

• Minimum Potential Energy ($\sf{U_{min}}$)=?

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Angular Velocity ($\sf{\omega}$):

$\large{\boxed{\sf \omega=\frac{k}{m}}}$

Therefore k= m${\omega^2}$

$\large\underline{\underline{\sf Maximum\:Kinetic\:Energy(K_{max})}}$

$\large{\boxed{\sf K_{max}=\frac{1}{2}m\omega^2A^2}}$

$\large\implies{\sf K_{max}=\frac{1}{2}×4×10^6×(0.01)^2 }$

$\large\implies{\sf K_{max}=200J}$

$\large\red{\boxed{\sf K_{max}=200J}}$

$\large\underline{\underline{\sf Minimum\: Potential\:Energy(U_{min})}}$

${\boxed{\sf U_{min}=Total\: Energy-Maximum\:Kinetic\: Energy}}$

$\large\implies{\sf U_{min}=240-200 }$

$\large\implies{\sf U_{min}=40J}$

$\large\red{\boxed{\sf U_{min}=40J }}$

Hence ,

Maximum Kinetic Energy is 200J and minimum Potential Energy is 40J

Answer 2

Answer:

That will be

$p.e. = \frac{1}{2} k {x}^{2} = \frac{1}{2} \times 4 \times {10}^{6} \times (0.01) {}^{2} = 200j$

So Maximum potential energy will be 200 J.

Maximum Kinetic Energy =200 J

40 J energy is of some other kind