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 x . Which of the following statement is true?

  • Option 1)

    K.E. is maximum when x = 0

  • Option 2)

    T.E. is zero when x = 0

  • Option 3)

    K.E. is maximum when x is maximum

  • Option 4)

    P.E. is maximum when x = 0

 

Answers (1)

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 x = 0

This is the correct option.

Option 2)

T.E. is zero when x = 0

This is an incorrect option.

Option 3)

K.E. is maximum when x is maximum

This is an incorrect option.

Option 4)

P.E. is maximum when x = 0

This is an incorrect option.

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