Question

A mass of 2.10 kg is attached to an oscillating spring, and the period of oscillation is 1.50s. Find the spring constant, and if the total energy is 4.55J find the amplitude of the oscillations.

Answer 1

Answer on Question #49803-Physics-Mechanics-Kinematics-Dynamics

A mass of $m = 2.10 \, kg$ is attached to an oscillating spring, and the period of oscillation is $T = 1.50s$. Find the spring constant, and if the total energy is $E = 4.55J$ find the amplitude of the oscillations.

Solution

The period of oscillation is

$$T = 2 \pi \sqrt {\frac {m}{k}},$$

where $k$ is the spring constant.

So, the spring constant is

$$k = m \left(\frac {T}{2 \pi}\right) ^ {- 2} = 2.10 \left(\frac {1.50}{2 \pi}\right) ^ {- 2} = 36.8 \frac {N}{m}.$$

The total energy is

$$E = \frac {k A ^ {2}}{2},$$

where $A$ is the amplitude of the oscillations.

Thus,

$$A = \sqrt {\frac {2 E}{k}} = \sqrt {\frac {2 \cdot 4.55}{36.8}} = 0.5 \, m.$$

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