# A body executes simple harmonic motion. The  potential energy (P.E.), the kinetic energy (K.E.) and total energy (T.E.) are measured as function of displacement . Which of the following statement is true? Option 1) K.E. is maximum when = 0 Option 2) T.E. is zero when = 0 Option 3) K.E. is maximum when is maximum Option 4) P.E. is maximum when = 0

As we learnt in

Kinetic energy in S.H.M. -

$K.E.= \frac{1}{2}mu^{2}\\ \: \: \: = \frac{1}{2}m\left ( A^{2} -x^{2}\right )\omega ^{2}$

- wherein

$K.E.= \frac{1}{2}K\left ( A^{2}-x^{2} \right ) \\K= m\omega ^{2}$

We know that

$K=\frac{1}{2}m\omega^{2}(A^{2}-x^{2})$

If x = 0

$K=\frac{1}{2}m\omega^{2}A^{2}=Maximum$

i.e. when x = 0 K.E. is maximum.

Correct option is 1.

Option 1)

K.E. is maximum when = 0

This is the correct option.

Option 2)

T.E. is zero when = 0

This is an incorrect option.

Option 3)

K.E. is maximum when is maximum

This is an incorrect option.

Option 4)

P.E. is maximum when = 0

This is an incorrect option.

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