Question

A force of \(6.4 \mathrm{~N}\) stretches a vertical spring by \(0.1 \mathrm{~m}\). The mass that must be suspended from the spring so that it oscillates with a time period of \(\pi / 4\) second. (1) \(\frac{\pi}{4} \mathrm{~kg}\) (2) \(\frac{4}{\pi} \mathrm{kg}\) (3) \(1 \mathrm{~kg}\) (4) \(10 \mathrm{~kg}\)

Short answer

1 kg

Step-by-step solution

Determine the spring constant

The force required to stretch the spring is given as 6.4 N and the extension caused is 0.1 m. The spring constant can be determined using Hooke's Law: \[ F = k x \]Rearranging for k gives us: \[ k = \frac{F}{x} = \frac{6.4 \text{ N}}{0.1 \text{ m}} = 64 \text{ N/m} \]

Find the mass required

Using the time period formula for a mass-spring system, we know: \[ T = 2\pi \sqrt{\frac{m}{k}} \]Given that the time period T is \( \frac{\pi}{4} \) seconds, we can substitute the known values and solve for m: \[ \frac{\pi}{4} = 2\pi \sqrt{\frac{m}{64}} \]First, simplify the equation: \[ \frac{1}{8} = \sqrt{\frac{m}{64}} \]Square both sides to remove the square root: \[ \frac{1}{64} = \frac{m}{64} \]Solve for m: \[ m = 1 \text{ kg} \]

Key concepts

Hooke's Law

Hooke's Law is fundamental in the study of springs and elasticity in physics. It describes the relationship between the force applied to a spring and the displacement caused by it. The law states that the force \(F\) required to compress or extend a spring is directly proportional to the displacement \(x\). The formula is given by \[ F = kx \], where \(k\) is the spring constant.The spring constant \(k\) is a measure of a spring's stiffness. The higher the value of \(k\), the stiffer the spring. Hooke’s Law is only valid within the elastic limit of a spring. If you stretch the spring beyond its elastic limit, it will not return to its original length.

Spring Constant

The spring constant \(k\) is crucial for understanding how different springs behave under various loads. It is measured in N/m (Newtons per meter). To determine \(k\), we can rearrange Hooke's Law equation to \[ k = \frac{F}{x} \]. This tells us that the spring constant equals the force applied divided by the displacement of the spring.In our example, a force of 6.4 N stretches the spring by 0.1 m. Using the formula, we find that the spring constant is \[ k = \frac{6.4}{0.1} = 64 \text{ N/m} \]. This high value indicates a relatively stiff spring.

Mass-Spring System

A mass-spring system is a classic example in physics that involves a mass attached to a spring. The system exhibits simple harmonic motion when displaced from its equilibrium position.The dynamics of a mass-spring system are described by the equation \[ F = ma \], where \(a\) is the acceleration. The restoring force provided by the spring follows Hooke's Law, \[ F = -kx \]. By combining Newton's second law with Hooke's Law, we can understand the oscillatory motion of the system. The mass-spring system is pivotal in studying oscillations and waves.

Oscillation Period

The oscillation period \(T\) of a mass-spring system is the time it takes for the system to complete one full cycle of motion. The period depends on the mass \(m\) and the spring constant \(k\) and is given by the formula \[ T = 2\pi \sqrt{\frac{m}{k}} \].In our example, we are given that the period \(T\) is \frac{\pi}{4}\ seconds. By substituting the known values into the formula, we derived the mass needed for this period: \[ \frac{\pi}{4} = 2\pi \sqrt{\frac{m}{64}} \]. Simplifying and solving for \(m\) gives us \[ m = 1 \text{ kg} \]. This understanding is essential for designing systems with specific oscillation characteristics.

Physics Equations

Physics equations form the backbone of problem-solving in topics like Hooke's Law and mass-spring systems. The ability to manipulate these equations is crucial. Here’s a quick recap:

• Hooke’s Law: \[ F = kx \]
• Spring Constant: \[ k = \frac{F}{x} \]
• Oscillation Period: \[ T = 2\pi \sqrt{\frac{m}{k}} \]

Understanding and applying these formulas correctly allows us to solve a wide range of physics problems. Always start by identifying known values, rearranging equations as needed, and performing careful calculations to arrive at the solution.