Block 2 of mass 1.90kg oscillates on the end of a spring in SHM with a period of

Question

Block 2 of mass 1.90kg oscillates on the end of a spring in SHM with a period of 12.00 ms. The position of the block is given by x = (1.60 cm) cos(omega t + pi/2). Block 1 of mass 3.80 kg slides toward block 2 with a velocity of magnitude 3.60 m/s, directed along the spring's length. The two blocks undergo a completely inelastic collision at time t = 3.00 ms. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision? Number 0.0305 Units m the tolerance is +/-2%

Explanation / Answer

from the given data

angular frequency,

w = 2*pi/T = 2*pi/(12*10^-3) = 523.6 rad/s

let k is the spring constant.

w = sqrt(k/m2)

==> k = w^2*m2

= 523.6^2*1.9

= 520898 N/m

we know, v = dx/dt

= 1.6*(-sin(w*t + pi/2))*w

= -1.6*523.6*10^-2*sin(523.6*t + pi/2)

= -8.38*sin(523.6*t + pi/2)

at t = 3 ms

v = -8.38*sin(523.6*3*10^-3 + pi/2)

= 0

Apply conservation of momentum

let v is the speed of both objects after the collision,

1.9*3.6 = (1.9 + 3.8)*v

==> v = 1.9*3.6/(1.9 + 3.8)

= 1.2 m/s

now use, Vmax = A*w

Vmax = A*sqrt(k/(m1+m2))

==> A = Vmax/sqrt(k/(m1+m2))

= 1.2/sqrt(520898/(3.8+1.9))

= 0.00397 m

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