A mass hung from a spring stretches the spring by 11 cm when the mass is resting at its equillibrium position. What will be the frequency of oscillation of this system if the mass is put in motion?
angFreq=sqrt(k/m)
k=mass*g/.11
2PI*freq=sqrt(9.8/.11)
solve for freq in hz
To calculate the frequency of oscillation of the system, we need to know the spring constant (k) and the mass (m).
The formula for the frequency (f) of an oscillating system is:
f = 1/(2π) * √(k/m)
Given that the spring stretches by 11 cm when the mass is at its equilibrium position, we can use this information to find the spring constant.
According to Hooke's Law, the force exerted by the spring (F) is proportional to the displacement (x):
F = -kx
In this case, when the mass is at its equilibrium position, there is no displacement and therefore no force applied (F = 0). So we can write:
0 = -k * 0
0 = 0
However, we know that when the mass is stretched 11 cm (or 0.11 m), it experiences a force equal to its weight:
F = mg
Where g is the acceleration due to gravity and m is the mass. Let's assume a value of g = 9.8 m/s^2.
Thus,
mg = kx
m * 9.8 = k * 0.11
We can rearrange this equation to solve for the spring constant (k):
k = (m * 9.8) / 0.11
Now we can substitute the known values for mass and spring constant into the formula for frequency:
f = 1/(2π) * √(k/m)
f = 1 / (2π) * √(((m * 9.8) / 0.11) / m)
f = 1 / (2π) * √((9.8 / 0.11))
f = 1 / (2π) * √(89.09)
f ≈ 0.164 Hz
Therefore, the frequency of oscillation of the system will be approximately 0.164 Hz.
To find the frequency of oscillation of the system, we can use the formula for the frequency of a mass-spring system, which is given by:
f = 1 / (2π) * √(k / m)
Where:
- f represents the frequency of oscillation
- k is the spring constant
- m is the mass
To find the frequency, we need to determine the spring constant. In this case, the spring stretches by 11 cm when the mass is resting at its equilibrium position.
The spring constant (k) can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Mathematically, it can be expressed as:
F = -k * x
Where:
- F is the force exerted by the spring
- x is the displacement from the equilibrium position
In this scenario, the spring is stretched by 11 cm, or 0.11 m, which is the displacement (x). The force exerted by the spring is equal to the weight of the mass, given by:
F = m * g
Where:
- m is the mass
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Setting these two equations equal to each other, we have:
m * g = -k * x
Rearranging the equation to solve for k:
k = -(m * g) / x
Now that we have the value of the spring constant, we can substitute it into the formula for the frequency of oscillation:
f = 1 / (2π) * √(k / m)
Substituting the value of k, we can calculate the frequency. Let's assume the mass is given in kilograms (kg).