Asked by s
A spring with a 4-kg mass has natural length 1 m and is maintained stretched to a length of
1.3m by a force of 24.3N. If the spring is compressed to a length of 0.8m and then released
with zero velocity, find the position of the mass at any time t.
1.3m by a force of 24.3N. If the spring is compressed to a length of 0.8m and then released
with zero velocity, find the position of the mass at any time t.
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Answered by
GPT-5 mini
Spring constant k = 24.3 N / 0.3 m = 81 N/m. For mass m = 4 kg the angular frequency is
omega = sqrt(k/m) = sqrt(81/4) = 4.5 rad/s.
Take x(t) as the displacement from the spring's natural length (positive for extension). Initially the length is 0.8 m so x(0) = 0.8 − 1 = −0.2 m and x'(0)=0. The SHM solution
x(t) = A cos(omega t) + B sin(omega t)
gives A = −0.2, B = 0, hence
x(t) = −0.2 cos(4.5 t) (meters).
If you want the actual spring length,
L(t) = 1 + x(t) = 1 − 0.2 cos(4.5 t) (meters).
(The period is T = 2π/4.5 ≈ 1.396 s.)
omega = sqrt(k/m) = sqrt(81/4) = 4.5 rad/s.
Take x(t) as the displacement from the spring's natural length (positive for extension). Initially the length is 0.8 m so x(0) = 0.8 − 1 = −0.2 m and x'(0)=0. The SHM solution
x(t) = A cos(omega t) + B sin(omega t)
gives A = −0.2, B = 0, hence
x(t) = −0.2 cos(4.5 t) (meters).
If you want the actual spring length,
L(t) = 1 + x(t) = 1 − 0.2 cos(4.5 t) (meters).
(The period is T = 2π/4.5 ≈ 1.396 s.)
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