91Ó°ÊÓ

In simple harmonic motion (SHM), angular frequency is a critical concept. For SHM, the position of an object can be described by the function \(x(t) = A \, \cos(\omega t + \phi)\). Here, \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase constant.
Angular frequency \(\omega\) measures how quickly an object oscillates back and forth. It’s related to the period \(T\), the time it takes for one complete cycle of motion, by the equation: \(\omega = \frac{2\pi}{T}\).
In our exercise, the period \(T\) is given as \(1.0 \, s\). Substituting into the formula, we get \(\omega = \frac{2\pi}{1.0 \, s} = 2\pi \, rad/s\). This tells us that the angular frequency of the piston's motion is \(2\pi \, rad/s\).
Understanding \(\omega\) helps us determine important factors like acceleration and velocity in SHM. These are crucial for predicting when a block will stay in contact with a moving piston.

The amplitude \(A\) in simple harmonic motion is the maximum distance the object moves from its equilibrium position. It’s a measure of how large the oscillations are.
In the given problem, to find the amplitude at which the block and piston will separate, we need to set the piston's maximum acceleration equal to the acceleration due to gravity, \(g = 9.8 \, m/s^2\). The maximum acceleration in SHM is given by \(a_{max} = A \omega^2\).
We already calculated \(\omega = 2\pi \, rad/s\). Setting \(A \omega^2 = g\), we get \(A (2\pi)^2 = 9.8\). Solving for \(A\):

• \(A \cdot 4\pi^2 = 9.8\)
• \(A = \frac{9.8}{4\pi^2} = \frac{9.8}{39.478} = 0.248 \, m = 24.8 \, cm\)

So, the block and piston will separate at an amplitude of 24.8 cm. This means that any motion larger than this will cause the block to leave the piston's surface.

Maximum acceleration \(a_{max}\) in SHM is a key concept, especially when ensuring continuous contact between two objects in motion, like the block and piston.
Maximum acceleration is calculated using \(a_{max} = A \omega^2\). This value tells us the greatest acceleration experienced by the object in SHM. For continuous contact, this maximum acceleration must not exceed the force due to gravity, \(g\).
In the exercise, we determined that the block and piston will separate if the maximum acceleration exceeds \(9.8 \, m/s^2\). For part (b), we need to find the maximum frequency \(f\) for a given amplitude \(A = 0.05 \, m\).
We use the formula \(a_{max} = A (2\pi f)^2 = g\) and solve for \(f\):

• \(0.05 (2\pi f)^2 = 9.8\)
• \((2\pi f)^2 = \frac{9.8}{0.05} = 196\)
• \(2\pi f = \sqrt{196} = 14\)
• \(f = \frac{14}{2\pi} = \frac{14}{6.283} = 2.23 \, Hz\)

So, the maximum frequency for continuous contact is about 2.23 Hz. This frequency ensures that the piston's motion doesn’t cause the block to lose contact.