Question

A partical exicute SHM with a time period 8second find the time in which 1/2 the total energy is potential energy

Answer 1

Answer:

x = A sin wt          describes SHM

v = w A cos w t     velocity

KE = 1/2 m v^2 = 1/2 m w^2 A^2 cos^2 w t

Max KE = 1/2 m w^2 A^2

KE = 1/4 m w^2 A^2    when KE = PE = 1/2 total energy

1/4 m w^2A^2 = 1/2 m w^2 A^2 cos^2 w t

1/2 = cos^2 w t

cos w t = 2^1/2 / 2

w t = 45 deg     when KE is 1/2 total energy

w t = pi / 4 = 45 deg

T = 8 =   1 / f  = 2 pi / w

So w = pi / 4     and w t = pi / 4    when KE is 1/2 total energy

So t must be 1 sec

Check: v = w A cos pi / 4     at 1 sec

v = w A * 2^1/2 / 2

v^2 = w^2 A^2 / 2

1/2 m v^2 = 1/4 m v^2 A^2  which is 1/2 the maximum KE and thus 1/2 the maximum PE  and this occurs at time t = 1