A block attached to a spring with unknown spring constant oscillates with a period of 6.0 s. What is the period if the mass is doubled?
The period is the same. If you do a proof with different equations, the mass of an object doesn't change its period. However, if the length of the spring was changed the period then would change.
When you do the proof, the mass ends up canceling out of the equation so it can be ignored.
To find the period of an oscillating system, we can use the equation:
T = 2π√(m/k)
where T is the period, m is the mass, and k is the spring constant.
In this case, the initial period is 6.0 s. We want to find the period when the mass is doubled. Let's denote the initial mass as m₁ and the final mass as m₂ = 2m₁.
Since the spring constant is unknown, we cannot directly determine its value. However, we can use the concept of the mass-spring system to relate the spring constant to the period.
From the equation T = 2π√(m/k), we can rewrite it as:
k = (4π²m) / T²
Since k is inversely proportional to the square of T, we can assume that k₁/T₁² = k₂/T₂², where k₁ and T₁ represent the initial spring constant and period, respectively, and k₂ and T₂ represent the final spring constant and period.
Let's substitute the values into the equation:
k₁/T₁² = k₂/T₂²
k₁/(6.0 s)² = k₂/T₂²
Now, let's double the mass: m₂ = 2m₁.
k₁/(6.0 s)² = k₂/(T₂₂)²
k₁/36.0 s² = k₂/T₂²
We can observe that k₁/36.0 s² is the ratio between the spring constant and the initial period squared, and k₂/T₂² is the ratio between the spring constant and the final period squared. Since the ratio remains constant, we can write:
k₁/36.0 s² = k₂/T₂²
Solving for T₂², we have:
T₂² = (k₁/36.0 s²) * T₂²
Now, let's substitute the final mass for m₂ into the equation:
T₂² = (k₁/36.0 s²) * T₂²
T₂² = (k₁/36.0 s²) * (2m₁)
Since m₂ = 2m₁:
T₂² = (k₁/36.0 s²) * (m₂)
The equation shows that the period squared is directly proportional to the mass. As the mass is doubled, the period squared will also double.
Therefore, the period when the mass is doubled will be √2 times the initial period. The initial period is 6.0 s, so the final period will be √2 * 6.0 s.
Final Period = √2 * 6.0 s ≈ 8.49 s
Hence, the period of the oscillation when the mass is doubled is approximately 8.49 seconds.