Discuss a simple harmonic motion in detail with your choice of examples. A massless spring (force constant = 3.60 N/cm) is cut into two equal parts and suspended separately to hold the mass M as shown in the figure. Find the mass of the block (M=?) if the system vibrates at a frequency of 2.87 Hz.

Simple harmonic motion (SHM) is a fascinating concept in physics that describes the oscillatory motion of an object around an equilibrium position. This type of motion is characterized by a restoring force that is directly proportional to the displacement from that equilibrium position, leading to a sinusoidal pattern in time. To illustrate this, let’s delve into the details of SHM and then tackle the specific problem involving a massless spring and a vibrating system.

Understanding Simple Harmonic Motion

In SHM, the motion can be described mathematically by the equation:

x(t) = A cos(ωt + φ)

Here, x(t) is the displacement from the equilibrium position, A is the amplitude (the maximum displacement), ω is the angular frequency, t is time, and φ is the phase constant. The key characteristics of SHM include:

• Restoring Force: The force that brings the object back to equilibrium is proportional to the displacement.
• Period and Frequency: The period (T) is the time taken for one complete cycle, while frequency (f) is the number of cycles per unit time, related by the equation f = 1/T.
• Energy Conservation: The total mechanical energy in SHM remains constant, oscillating between potential and kinetic energy.

Examples of Simple Harmonic Motion

Common examples of SHM include:

• Pendulum: A swinging pendulum exhibits SHM when the angle of swing is small.
• Mass on a Spring: A mass attached to a spring will oscillate when displaced from its equilibrium position.

Analyzing the Spring-Mass System

Now, let’s focus on the specific scenario you provided, where a massless spring with a force constant of 3.60 N/cm is cut into two equal parts. Each part will have a new spring constant, and we need to find the mass (M) that allows the system to vibrate at a frequency of 2.87 Hz.

Calculating the New Spring Constant

When the spring is cut into two equal parts, the spring constant of each half increases. The relationship for the spring constant when a spring is cut is given by:

k' = 2k

Where k' is the new spring constant and k is the original spring constant. Given that the original spring constant is 3.60 N/cm, we convert this to N/m for consistency:

3.60 N/cm = 360 N/m

Thus, the new spring constant for each half is:

k' = 2 × 360 N/m = 720 N/m

Finding the Mass Using Frequency

The frequency of a mass-spring system in SHM is given by the formula:

f = (1/2π) √(k/m)

Rearranging this formula to solve for mass (m), we have:

m = k/(4π²f²)

Substituting the known values:

• k = 720 N/m
• f = 2.87 Hz

Now, we can calculate:

m = 720 / (4π²(2.87)²)

Calculating the denominator:

4π²(2.87)² ≈ 4 × 9.87 × 8.22 ≈ 324.5

Now substituting back to find the mass:

m ≈ 720 / 324.5 ≈ 2.22 kg

Final Thoughts

In summary, the mass that allows the system to vibrate at a frequency of 2.87 Hz is approximately 2.22 kg. This example not only illustrates the principles of simple harmonic motion but also shows how to apply mathematical concepts to real-world scenarios. Understanding these principles is essential for analyzing various physical systems in mechanics.