A 2.0kg block hang from a vertical spring causes it to stretch by 200cm. If the 2.0kg block is replaced by a 0.50kg mass and the spring is stretched and released, what are the frequency and period of the oscillations?
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m₁g=kx
k= m₁g/x
ω=sqrt(k/m₂)=… (rad/s)
f=ω/2π=…(Hz)
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The frequency and period of the oscillations can be determined using the following equations:
1. For the initial situation with a 2.0kg block hanging from the spring:
From Hooke's Law, we know that F = kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring.
The weight of the block is given by m₁g, where m₁ = 2.0kg and g = 9.8m/s².
Setting F = m₁g and x = 200cm = 2.0m, we can solve for the spring constant:
k = F/x = m₁g/x = (2.0kg)(9.8m/s²)/(2.0m) = 19.6 N/m.
2. For the situation with a 0.50kg mass replacing the 2.0kg block:
We can find the new equilibrium position by equating the weight of the mass to the force exerted by the spring:
m₂g = kx₀,
where m₂ = 0.50kg, g = 9.8m/s², x₀ is the new equilibrium position.
Solving for x₀:
x₀ = m₂g/k = (0.50kg)(9.8m/s²)/(19.6 N/m) = 0.25m.
Since the spring was initially stretched by 2.0m, the extension of the spring when the 0.50kg mass replaces the 2.0kg block is:
Δx = x - x₀ = 2.0m - 0.25m = 1.75m.
3. The frequency of the oscillations can be found using the formula:
ω = sqrt(k/m₂),
where m₂ = 0.50kg and k = 19.6 N/m.
ω = sqrt(19.6 N/m / 0.50kg) = sqrt(39.2 N/kg) ≈ 6.26 rad/s.
4. Finally, the period of the oscillations is given by:
T = 2π/ω = 2π / 6.26 rad/s ≈ 1.00 s.
So, the frequency of the oscillations is approximately 6.26 Hz and the period is approximately 1.00 s.
To solve this problem, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement (stretch or compression) of the spring from its equilibrium position.
Let's start by finding the spring constant, k.
The spring constant represents the stiffness of the spring and is given by the formula:
k = F / x
Where:
k is the spring constant
F is the force applied to the spring (weight of the block in this case)
x is the displacement of the spring (stretch or compression)
Given that the 2.0 kg block stretches the spring by 200 cm (which is equal to 2.0 m), we can calculate the force F applied to the spring:
F = m * g
Where:
F is the force applied to the spring
m is the mass of the block
g is the acceleration due to gravity (9.8 m/s^2)
F = 2.0 kg * 9.8 m/s^2
F = 19.6 N
Now we can calculate the spring constant, k:
k = F / x
k = 19.6 N / 2.0 m
k = 9.8 N/m
Now, let's calculate the new period and frequency when the 0.50 kg mass is used.
To calculate the period (T) of oscillation:
T = 2π * √(m / k)
Where:
T is the period of oscillation
m is the mass of the object attached to the spring
k is the spring constant
T = 2π * √(0.50 kg / 9.8 N/m)
T ≈ 2.83 s
To calculate the frequency (f) of oscillation:
f = 1 / T
f = 1 / 2.83 s
f ≈ 0.354 Hz
So, the period of the oscillation is approximately 2.83 seconds, and the frequency is approximately 0.354 Hz.