At what time t1 does the block come back to its original equilibrium position (x=0) for the first time? Express your answer in terms of some or all of the variables: A, k, and m.

 t_1 = 0.5*pi*sqrt( m / k )    

Explanation:

Given:

- The block of mass m undergoes simple harmonic motion. With the displacement of x from mean position is given by:

                                     x(t) = A*cos(w*t)

Find:

- At what time t1 does the block come back to its original equilibrium position (x=0) for the first time?

Solution:

- The first time the block moves from maximum position to its mean position constitutes of 1/4 th of one complete cycle. So, the required time t_1 is:

                                            t_1 = 0.25*T

- Where, T : Time period of SHM.

- The time period for SHM is given by:

                                         T = 2*pi*sqrt ( m / k )

Hence,

                                   t_1 = 0.25 * 2 * pi * sqrt( m / k )

                                   t_1 = 0.5*pi * sqrt( m / k )    

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

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