It will take 1.47 seconds for the mass to reach the new equilibrium position.
A pendulum is in equilibrium when its potential energy equals its kinetic energy. In other words, the kinetic energy is zero and the potential energy is maximum when the pendulum is at rest and the bob is at the lowest point of its arc.
We can use the following equation to calculate the period of oscillation of a mass on a spring:
T = 2π√(m/k)
Where T is the period, m is the mass, and k is the spring constant.
We can use the equation to calculate the spring constant, k, as follows:
k = m/x²
Where x is the initial displacement of the spring.
In this case, m = 1.65 kg and x = 0.270 m.
Therefore, the spring constant, k, is:
k = 1.65/0.27² = 17.59 N/m
We can now use the first equation to calculate the period of oscillation:
T = 2π√(1.65/17.59) = 1.56 s
Since the spring has been stretched an additional 0.130 m, we can calculate the new period of oscillation as follows:
T = 2π√(1.65/19.72) = 1.47 s
Therefore, it will take 1.47 seconds for the mass to reach the new equilibrium position.
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Complete question:
A 1.65 kg mass stretches a vertical spring 0.270 m. If the spring is stretched an additional 0. 130 m and released, how long does it take to reach the (new) equilibrium position again?.
