Question

A spring with a mass of 4 kg has natural length 0.5 m. A force of 25.6 N is required to maintain it stretched to a length of 0.7 m. If the spring is stretched to a length of 0.7 m and then released with initial velocity 0, find the position of the mass at any time.

Answer 1

Answer on Question#49746 - Physics - Mechanics - Kinematics - Dynamics

A spring with a mass of $4\mathrm{kg}$ has natural length $0.5\mathrm{m}$ . A force of $25.6\mathrm{N}$ is required to maintain it stretched to a length of $0.7\mathrm{m}$ . If the spring is stretched to a length of $0.7\mathrm{m}$ and then released with initial velocity 0, find the position of the mass at any time.

Solution:

Since the force of $25.6\mathrm{N}$ is required to maintain the spring stretched by the length of $0.2\mathrm{m}$ , the stiffness of the string has the following value

$$k = \frac {2 5 . 6 \mathrm {N}}{0 . 2 \mathrm {m}} = 1 2 8 \frac {\mathrm {N}}{\mathrm {m}}$$

According to the 2 Newton's law the equation of motion of the mass (under the restoring force) can be written as follows

$$M \ddot {x} = - k (x - 0, 5 \mathrm {m})$$

where $M = 4\mathrm{kg}$ and $x$ is the position of the mass. The solution of this equation is

$$x = 0, 5 \mathrm {m} + A \sin \frac {k}{M} t + B \cos \frac {k}{M} t$$

where $A$ and $B$ are some constants. We can determine them from the initial conditions, which are

$$x (0) = 0. 7 \mathrm {m}, \dot {x} (0) = 0 \frac {\mathrm {m}}{\mathrm {s}}$$

Using the first condition we obtain

$$0, 7 \mathrm {m} = 0, 5 \mathrm {m} + B$$

It gives us the value of $B$ :

$$B = 0, 2 \mathrm {m}$$

Taking the derivative of (1) we obtain

$$\dot{x} = A \frac{k}{M} \cos \frac{k}{M} t - B \frac{k}{M} \sin \frac{k}{M} t$$

Using second condition we obtain

$$\dot{x}(0) = A \frac{k}{M} = 0$$

It gives us the value of $A$:

$$A = 0$$

Therefore, the position of the mass at any time is given by

$$x = 0,5\,\mathrm{m} + 0,2\,\mathrm{m} \cdot \cos \frac{128\,\frac{\mathrm{N}}{\mathrm{m}}}{4\,\mathrm{kg}} t = 0,5\,\mathrm{m} + 0,2\,\mathrm{m} \cdot \cos(32\,\mathrm{s}^{-1} \cdot t)$$

Answer: $x = 0,5\,\mathrm{m} + 0,2\,\mathrm{m} \cdot \cos(32\,\mathrm{s}^{-1} \cdot t)$.

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