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A block of mass m is hung from a vertical spring and allowed to reach equilibrium at rest. The spring constant is k=0.414N/m. If the block is then pulled down an additional distance of 0.668 m and released from rest, what is its period of oscillation?

User Beingnin
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2 Answers

3 votes

Final Answer:

The period of oscillation for the block pulled down an additional distance of 0.668 m and released from rest, is approximately T = 2π√(m/k), where m is the mass of the block and k is the spring constant.

Step-by-step explanation:

When the block is pulled down and released, it undergoes simple harmonic motion (SHM). The formula for the period (T) of an object undergoing SHM is given by T = 2π√(m/k), where m is the mass and k is the spring constant. In this case, the additional distance of 0.668 m doesn't affect the equilibrium position, but it does impact the amplitude of oscillation.

The key insight here is that the period of oscillation in simple harmonic motion depends only on the mass and the spring constant, not on the amplitude. Thus, the given period is determined solely by the initial conditions of the system, specifically the mass of the block and the spring constant. The formula T = 2π√(m/k) encapsulates this relationship, providing a straightforward way to calculate the period.

In summary, the period of oscillation is determined by the interplay between the mass of the block and the spring constant, as given by the formula T = 2π√(m/k). The additional distance of 0.668 m, while affecting the amplitude, does not alter the period of oscillation in this scenario.

User Abguy
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7.2k points
5 votes

Final answer:

The period of oscillation for the mass-spring system with a spring constant of 0.414 N/m is approximately 9.79 seconds.

Step-by-step explanation:

When a block of mass m is hung from a vertical spring and then pulled down an additional distance and released, it undergoes simple harmonic motion (SHM).

The period of oscillation T of a mass-spring system in SHM is given by the formula T = 2π√(m/k), where m is the mass and k is the spring constant.

Given that the spring constant k is 0.414 N/m, we can find the period without needing the mass of the block, since it cancels out in the formula.

The period of oscillation is calculated as:

T = 2π√(m/0.414 N/m)

T = 2π√(1/0.414)

T ≈ 2π√(2.4155)

T ≈ 2π(1.5542)

T ≈ 9.7891 seconds

The period of oscillation of the block attached to the spring is approximately 9.79 seconds.

User Gangesh
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