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 determine the distance of the small mass from equilibrium position and the angular frequency at resonance, we'll need to understand the concept of Hooke's Law for springs and the equations governing simple harmonic motion.
1. Distance of the small mass from equilibrium position:
According to Hooke's Law, the force exerted by a spring is directly proportional to its displacement from equilibrium. The formula is given by:
F = -kx
where F is the force applied on the spring, k is the spring constant, and x is the displacement from equilibrium.
In this case, a force of 5*10^-4 N is applied, which causes the spring to expand by 5*10^-2 m. By rearranging the formula and plugging in the values, we can solve for the spring constant:
k = -F / x
k = -(5*10^-4 N) / (5*10^-2 m)
k = -0.01 N/m
Now, since a small mass of 0.01 kg is hung at the bottom end, we can find the acceleration caused by this force:
F = m * a
-0.01 N/m = 0.01 kg * a
a = -1 m/s^2 (negative sign indicates it is in the opposite direction to the applied force)
The distance of the small mass from the equilibrium position when in equilibrium is given by:
x_eq = F / k
x_eq = (0.01 kg * -1 m/s^2) / (-0.01 N/m)
x_eq = 1 m
Therefore, the small mass is 1 meter from the equilibrium position when in equilibrium.
2. Angular frequency at resonance:
The angular frequency (ω) at resonance can be determined using the formula:
ω = √(k / m)
where k is the spring constant and m is the mass hung on the spring.
In this case, k = -0.01 N/m and m = 0.01 kg, so:
ω = √((-0.01 N/m) / (0.01 kg))
ω = √(-1 rad/s^2)
ω = undefined
Since the angular frequency is undefined, it indicates that the system does not exhibit resonance under these conditions.
Therefore, the distance of the small mass from the equilibrium position is 1 meter, and the angular frequency at resonance is undefined in this case.