The total energy of particle is sum of kinetic & potential energy. Potential energy is frame dependent, so it can be negative in some cases. Thus, value of kinetic energy can exceed total energy.

The time period of given motion is T. It will go from centre to -A in T/4 secs. Then back to centre at 2T/4 secs. And n towards right extreme, A, at 3T/4 secs. At this extreme position its velocity will be zero.

When block is displaced by x in one direction, let?s say left side, all springs will exert a force on block towards right. The two springs will be compressed by length x and right one will be extended by length x.
Net force on block = kx + 2kx + (k/2)x = 3.5kx.
Displacement is towards left and force is towards right, so we write F = -3.5kx.
F = ma = -3.5kx.
&re4; a = -(3.5k/m)x.
We know that in SHM a = -w²x.
&re4; w = √(3.5k/m).
&re4; T = 2π/w = 2π √(m/3.5k) = 2π √(2m/7k) s.

The particle has zero acceleration at t=0, so it starts from mean position. Then acceleration becomes positive which implies that particle is moving towards -A ( negative extreme). The equation of acceleration can be given by: a = 8sinwt.
The time period is 2s,
so w = 2π/2 = πs^-1.
&re4; a = 8sin(πt).
&re4; v = -8/π cos(πt)
&re4; x = -8/pi²sin(πt) = -0.81sin(πt).

Their time period will be same as length is same. So time to reach mean position from ir respective extremes will be same. Therefore y will reach mean position at same time. They both have different potential energies at start, so ir maximum speeds will be different. Anor way to look at this is that speed of 2nd pendulum should be more as it has to cover greater arc length in same amount of time.

w = 2π/T = π.
Maximum kinetic energy = 1/2mA²w², which will be at x=0.
Total energy will be constant,
refore energy at x=0 is same as energy at x = +1.
Thus, energy = 1/2mA²w² – 5
= (0.5*2*1*9.87) – 5
= 4.87J.

The motion will repeat itself after ball hits ground every time. The time taken for ball to reach topmost position is t = u/g.
This will also be time to reach ground.
Thus, total time = 2u/g.
This will be time period.
Frequency = 1/T = g/2u
= 10/5 =2Hz.

x = cos²wt ? sin²wt = cos 2wt.
Thus, time period T = 2π/2w = π/w.
Also amplitude of motion is 1m.

Let displacement of a particle be given by: x = Asin(wt).
Velocity will be = Awsin(wt + π/2)
acceleration will be = -Aw²sin(wt) = Aw²sin(wt + π).
Thus, phase difference is π/2.

When block is released, spring force will act towards mean position and friction force will act against velocity. The work done by se forces should be equal to potential energy of system in compressed state. &re4; Work done in dx displacement = (?mg – kx)dx.
On integrating this, with dx varying from 0 to 0.1, we get
Work = (μmgx – kx√/2)₀^0.1= 0.5μ – 0.015.
This should be equal to initial potential energy of system.
0.5μ – 0.015 = 1/2kA√
&re4; 0.5μ – 0.015 = 1/2 * 3 * 0.01
&re4; 0.5μ – 0.015 = 0.015
&re4; 0.5μ = 0.03 OR μ = 0.06.