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 use Hooke's Law. Hooke's Law states that the force required to extend or compress a spring is directly proportional to the distance it is stretched or compressed.
The equation for Hooke's Law is given by:
F = k * x
Where:
F is the force applied on the spring,
k is the spring constant, and
x is the distance the spring is stretched or compressed.
In this case, the force applied on the spring is given as 5 * 10^(-4) N, and the distance the spring is expanded by is 5 * 10^(-2) m.
Using the equation, we can rearrange it to solve for the spring constant, k:
k = F / x
Plugging in the values, we get:
k = (5 * 10^(-4) N) / (5 * 10^(-2) m)
k = 0.01 N/m
Now, to find the distance of the mass from the equilibrium position, we can use the equation for simple harmonic motion (SHM):
x = A * cos(ωt)
Where:
x is the distance from the equilibrium position,
A is the amplitude (maximum distance from the equilibrium position), and
ω is the angular frequency.
In this case, the amplitude is not given, but we know that the force function is given as f(t) = 20 * cos(ωt).
At equilibrium, the force exerted by the spring is equal to the force applied at the top end. So:
k * x = f(t)
Substituting in the known values, we get:
(0.01 N/m) * x = 20 * cos(ωt)
Simplifying, we have:
x = (20 / 0.01) * cos(ωt)
x = 2000 * cos(ωt)
Since we want to find x when the small mass is at equilibrium, we can set ωt = π/2. This is because the cosine function is maximum at ωt = π/2.
x = 2000 * cos(π/2)
x = 2000 * 0
x = 0
Therefore, at equilibrium, the distance of the small mass from the equilibrium position is 0.
Now, to find the angular frequency at resonance, we need to know the relationship between the angular frequency (ω) and the natural frequency (ω₀) of the system.
In simple harmonic motion, the angular frequency (ω) is related to the natural frequency (ω₀) by the equation:
ω = ω₀ * √(1 - β²)
Where:
β = damping constant (assumed to be zero for an undamped system).
At resonance, the natural frequency (ω₀) is equal to the angular frequency (ω), resulting in maximum amplitude and energy transfer.
In this case, since the damping constant (β) is not given, we assume it to be zero.
Therefore, the angular frequency (ω) at resonance is equal to the natural frequency (ω₀), which can be calculated using the formula:
ω₀ = √(k / m)
Where:
k is the spring constant (0.01 N/m), and
m is the mass (0.01 kg).
Plugging in the values, we get:
ω₀ = √(0.01 N/m / 0.01 kg)
ω₀ = √(1)
ω₀ = 1 rad/s
Thus, the angular frequency at resonance is 1 rad/s.