A particle moves along the x axis. It is initially at the position 0.310 m, movi

Question

A particle moves along the x axis. It is initially at the position 0.310 m, moving with velocity 0.150 m/s and acceleration -0.260 m/s2. Suppose it moves with constant acceleration for 6.00 s.

We take the same particle and give it the same initial conditions as before. Instead of having a constant acceleration, it oscillates in simple harmonic motion for 6.00 s around the equilibrium position x = 0.

A particle moves along the x axis. It is initially at the position 0.310 m, moving with velocity 0.150 m/s and acceleration -0.260 m/s2. Suppose it moves with constant acceleration for 6.00 s. Find the position of the particle after this time. Find its velocity at the end of this time interval. We take the same particle and give it the same initial conditions as before. Instead of having a constant acceleration, it oscillates in simple harmonic motion for 6.00 s around the equilibrium position x = 0. Find the angular frequency of the oscillation. Hint: in SHM, a is proportional to x. Find the amplitude of the oscillation. Hint: use conservation of energy. Find its phase constant theta 0 if cosine is used for the equation of motion. Hint: when taking the inverse of a trig function, there are always two angles but your calculator will tell you only one and you must decide which of the two angles you need.

Explanation / Answer

D) Let us assume that the block is suspended from a spring of spring constant k

at extreme point total energy of system will be 1/2kx2 (since velocity is zero,no kinetic energy is present)

k= mw2

total energy at current position is

spring energy + kinetic energy

= 1/2 k*(0.310)2 + 1/2 m (0.15)2

= 1/2 * m * (0.915)2(0.310)2 + 1/2 m (0.15)2

total energy at extreme point is

spring energy

= 1/2 k (amplitude)2

= 1/2 m (0.915)2(amp)2

equating them due to conservation of energy

amp2 = (0.310)2 + (0.15/0.915)2

amplitude = 0.35

E) the motion follows the given equation

x = A cos (wt+ pi )

at current position time is t = 0 and x = 0.31

substituting in equation

0.31 = 0.35 cos (pi)

cos(pi) = 0.885

pi= 27.6 deg

F) for position after 6s,substituing 6 in the equation

x = a cos (wt + 27.6)

= 0.35 cos (0.915 * 6 + 27.6)

= 0.35 * 0.837

=0.293

G)

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