Question

The figure shows a mass \(m_{2}=20.0\) g resting on top of a mass \(m_{1}=20.0 \mathrm{~g}\) which is attached to a spring with \(k=10.0 \mathrm{~N} / \mathrm{m}\) The coefficient of static friction between the two masses is 0.600 . The masses are oscillating with simple harmonic motion on a frictionless surface. What is the maximum amplitude the oscillation can have without \(m_{2}\) slipping off \(m_{1} ?\)

Short answer

Answer: The maximum amplitude of the oscillation without the upper mass slipping off the lower mass is 0.011772 m.

Step-by-step solution

Determine the maximum friction force between the masses

To find the maximum friction force between the two masses, we can use the formula: \(max\_friction\_force = μ * normal\_force\) where \(μ = 0.600\) (the coefficient of static friction) and \(normal\_force = m_{2}g\), where \(g\) is the gravitational constant, approximately equal to \(9.81 m/s^2\). \(max\_friction\_force = 0.600 * (20.0 * 10^{-3}) * 9.81\)

Calculate the maximum friction force between the masses

Now, we can calculate the maximum friction force between the masses: \(max\_friction\_force = 0.600 * 0.020 * 9.81 N = 0.11772 N\)

Determine the force exerted by the spring at a certain amplitude

We know that the force exerted by the spring (\(F_{spring}\)) is given by Hooke's law: \(F_{spring} = k * x\) where \(k = 10.0 N/m\) (spring constant) and \(x\) is the displacement (or amplitude), which we are trying to find. For the upper mass not to slip off, \(F_{spring} = max\_friction\_force\)

Solve for the maximum amplitude

Now we can set the force exerted by the spring equal to the maximum friction force and solve for the maximum amplitude, \(x\): \(10.0 * x = 0.11772\) \(x = \frac{0.11772}{10.0}\)

Calculate the maximum amplitude

Finally, we can calculate the maximum amplitude: \(x = \frac{0.11772}{10.0} = 0.011772 m\) The maximum amplitude the oscillation can have without \(m_{2}\) slipping off \(m_{1}\) is 0.011772 m.

Key concepts

Static Friction

Static friction is the force that prevents two surfaces from sliding past each other. It acts in response to an applied force and will increase to a maximum value before motion occurs. Until this threshold is reached, static friction balances the applied forces, keeping the objects stationary relative to each other.

• Static friction is crucial in scenarios where objects are resting on a moving surface, like in this problem where mass \(m_{2}\) rests on mass \(m_{1}\).
• The equation for static friction is given by \(f_{s} = \mu \cdot N\), where \(\mu\) is the coefficient of static friction and \(N\) is the normal force.
• The coefficient \(\mu\) is a measure of how "grippy" the two surfaces are. A higher value means more friction.

In this problem, the normal force is due to gravity acting on mass \(m_{2}\), and is calculated as \(m_{2} \cdot g\), where \(g\) is the acceleration due to gravity. Thus, the maximum static friction force is the product of this normal force and the coefficient of static friction. If the force trying to move \(m_{2}\) surpasses this static friction, \(m_{2}\) will begin to slide off \(m_{1}\).

Spring Force

Spring force is a restoring force that pushes or pulls back to its equilibrium position once displaced. According to Hooke's Law, this force is directly proportional to the displacement from the spring's resting length, represented by the equation \(F_{spring} = k \cdot x\).

• The constant \(k\) in the equation is known as the spring constant, which measures the stiffness of the spring. A larger \(k\) means a stiffer spring.
• The variable \(x\) refers to the displacement or the amount the spring is stretched or compressed from its natural, resting state.

In our scenario, the spring force pulls back with increasing intensity as the displacement grows. This force helps to maintain the masses in oscillation and can be considered stable, if it does not exceed the static friction's maximum force. When the spring force equals the maximum static friction, the system is at its edge before slipping would occur, thus defining the maximum safe amplitude for the oscillation.

Amplitude Calculation

Amplitude in simple harmonic motion represents the maximum distance from the equilibrium position. It is a critical measure in oscillatory motion that determines the extent of swing in each cycle.To find the maximum amplitude where the mass \(m_{2}\) does not slip off the mass \(m_{1}\), we equate the maximum frictional force to the force exerted by the spring:1. Identify the maximum static friction force, which acts as the threshold for slipping.2. Use Hooke's Law to relate this friction force to the spring's displacement.3. Solve for the amplitude \(x\) where \(F_{spring} = k \cdot x = \text{max static friction}\).For this problem, we set \(k \cdot x = 0.11772 \) N (the calculated maximum static friction force). Solving the equation \(10 \cdot x = 0.11772\), we find the maximum amplitude \(x\):\[ x = \frac{0.11772}{10} = 0.011772 \text{ m} \]This means that the maximum amplitude of oscillation is 0.011772 meters, beyond which the top mass would no longer stay stationary relative to the bottom mass.