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 its equilibrium position when a force is applied at the top end, we can use Hooke's Law and the equation for simple harmonic motion.
1. Hooke's Law states that the force applied on a spring is proportional to the extension or compression of the spring. The formula is F = k * x, where F is the force applied, k is the spring constant, and x is the displacement from the equilibrium position.
2. In this case, the spring is extended by 5 * 10^-2 m when a force of 5 * 10^-4 N is applied. We can calculate the spring constant using the formula k = F / x.
k = (5 * 10^-4 N) / (5 * 10^-2 m)
= 0.01 N/m
3. Since a force F(t) = 20 cos(wt) is applied at the top end, the equation for the motion of the small mass is given by:
m * a = -k * x + F(t)
Here, m is the mass (0.01 kg), a is the acceleration, x is the displacement, k is the spring constant, and F(t) is the forcing function.
4. Since the system is in equilibrium, the acceleration is zero. Therefore, we have:
0 = -k * x + F(t)
Plugging in the values, we get:
0 = (-0.01 N/m) * x + 20 cos(wt)
5. Rearranging the equation, we find:
x = (20 cos(wt)) / (0.01 N/m)
Simplifying further:
x = 2000 cos(wt) m
This equation gives us the distance of the small mass from the equilibrium position as a function of time.
To find the angular frequency at resonance, we can use the formula:
w = sqrt(k / m)
Plugging in the values we calculated earlier:
w = sqrt(0.01 N/m / 0.01 kg)
= 1 rad/s
Therefore, the distance of the small mass from the equilibrium position is given by x = 2000 cos(wt) m, and the angular frequency at resonance is w = 1 rad/s.
To find the distance of the small mass from the equilibrium position, we need to calculate the displacement caused by the force applied.
Given:
Force applied on the spring = 5 * 10^-4 N
Expansion of the spring = 5 * 10^-2 m
Mass hanging at the bottom end = 0.01 kg
Force applied for oscillation = f(t) = 20cos(wt)
Step 1: Calculate the spring constant (k)
The spring constant (k) represents the stiffness of the spring. It can be calculated using Hooke's Law: F = k * x, where F is the force and x is the displacement.
In this case, using the given values:
5 * 10^-4 N = k * 5 * 10^-2 m
Dividing both sides by 5 * 10^-2 m:
k = (5 * 10^-4 N) / (5 * 10^-2 m)
k = 10 N/m
Step 2: Calculate the angular frequency (w) of the oscillation
The angular frequency (w) can be calculated using the formula: w = sqrt(k / m), where k is the spring constant and m is the mass.
In this case, using the given values:
m = 0.01 kg
k = 10 N/m
w = sqrt(10 N/m / 0.01 kg)
w = sqrt(1000)
w = 31.62 rad/s (rounded to two decimal places)
Step 3: Find the distance of the small mass from the equilibrium position
To find the distance of the small mass from the equilibrium position, we need to consider the force applied for oscillation and calculate the displacement caused by it.
The force applied for oscillation is given as f(t) = 20cos(wt).
Let's assume that the distance of the small mass from the equilibrium position is d meters.
Using Hooke's Law again, we can write the equation for the force on the small mass:
F = k * d
Comparing this with the given force f(t), we get:
k * d = 20cos(wt)
Substituting the value of k from Step 1 and rearranging the equation, we find:
d = (20cos(wt)) / k
Substituting the given values:
d = (20cos(wt)) / 10
d = 2cos(wt)
Therefore, the distance of the small mass from the equilibrium position is given by the equation d = 2cos(wt).