when a force (5*10^-4) is applied on a spring then it get expand by 5*10^-2.a small mass(0.01kg) is hanged at bottm end.after coming in equilibrium a force [f(t)=20cos(wt)] is applied at the top end for oscillation.what is the distance of small(0.01) from equilibrium position?what is angular frequency at resonancy?
To find the distance of the small mass (0.01 kg) from the equilibrium position, we can first calculate the spring constant (k) using the formula:
k = F / x
where F is the applied force (5 * 10^-4 N) and x is the expansion of the spring (5 * 10^-2 m).
k = (5 * 10^-4 N) / (5 * 10^-2 m)
k = 0.01 N/m
Next, we can calculate the restoring force of the spring (Fr) using Hooke's Law:
Fr = -k * x
where x is the displacement of the mass from the equilibrium position. Since the force applied (f(t)) is in simple harmonic motion, the restoring force (Fr) will be equal to the applied force:
Fr = f(t)
Therefore, we have:
-k * x = 20 * cos(ωt)
Substituting the value of k and rearranging the equation, we get:
(0.01 N/m) * x = -20 * cos(ωt)
Now, at equilibrium, the net force on the mass will be zero. Since the mass (m) is 0.01 kg, we can calculate the acceleration (a) using Newton's second law:
a = Fnet / m
At equilibrium, Fnet is zero, so the acceleration becomes zero as well. This means that the restoring force (Fr) and the applied force (f(t)) must cancel each other out:
Fr = f(t)
Substituting the value of Fr, we have:
(0.01 N/m) * x = -20 * cos(ωt)
Since we are looking for the distance of the small mass from the equilibrium position, we can solve the equation for x:
x = -(20 * cos(ωt)) / (0.01 N/m)
The angular frequency (ω) at resonance can be determined by the equation:
ω = √(k / m)
Substituting the values of k (0.01 N/m) and m (0.01 kg):
ω = √(0.01 N/m / 0.01 kg)
ω = √(1 N/m)
ω = 1 rad/s
Therefore, the distance of the small mass from the equilibrium position can be found using the equation x = -(20 * cos(ωt)) / (0.01 N/m), and the angular frequency at resonance is 1 rad/s.