Question

at what distance form the mean position is the kinetic energy of the particle performing S H M of amplitude 8 cm three times its potential energy​

Answer 1

The potential energy of a particle executing SHM at an amplitude $\sf{y}$ (say) is given by,

$\longrightarrow\sf{U(y)=\dfrac{1}{2}ky^2}$

where $\sf{k}$ is the force constant.

The potential energy of the particle executing SHM at its maximum amplitude $\sf{a}$ (say) is maximum.

$\longrightarrow\sf{U_{max}=U(a)}$

$\longrightarrow\sf{U_{max}=\dfrac{1}{2}ka^2}$

Maximum value is because the particle experiences no velocity at extreme positions and hence no kinetic energy.

$\longrightarrow\sf{K(a)=0}$

At amplitude $\sf{y,}$ the kinetic energy experienced is given by total mechanical energy conservation.

$\longrightarrow\sf{K(y)+U(y)=K(a)+U(a)}$

$\longrightarrow\sf{K(y)+\dfrac{1}{2}ky^2=\dfrac{1}{2}ka^2}$

$\longrightarrow\sf{K(y)=\dfrac{1}{2}k(a^2-y^2)}$

========================================

By the question the kinetic energy of the particle at a particular amplitude is thrice its potential energy, i.e.,

$\longrightarrow\sf{K(y)=3\cdot U(y)}$

$\longrightarrow\sf{\dfrac{1}{2}k(a^2-y^2)=3\cdot\dfrac{1}{2}\,ky^2}$

$\longrightarrow\sf{a^2-y^2=3y^2}$

$\longrightarrow\sf{4y^2=a^2}$

$\longrightarrow\sf{y^2=\dfrac{a^2}{4}}$

$\longrightarrow\sf{\underline{\underline{y=\pm\,\dfrac{a}{2}}}}$

I.e., the condition is satisfied if the particle reaches half of its maximum amplitude.