Askiitians Tutor Team
Last Activity: 5 Months ago
To solve the problem of a mass oscillating harmonically, we can use some fundamental principles from physics. Let's break it down step by step, focusing on the period of oscillation and the maximum kinetic energy of the mass.
1. Finding the Period of Oscillation
The period of oscillation for a mass-spring system can be calculated using the formula:
T = 2π√(m/k)
Where:
- T is the period of oscillation.
- m is the mass (in kg).
- k is the spring constant (in N/m).
In this case, we have:
Now, substituting the values into the formula:
T = 2π√(0.5 kg / 10 N/m)
Calculating the value inside the square root:
T = 2π√(0.05)
Now, calculating √(0.05):
√(0.05) ≈ 0.2236
Thus, we have:
T ≈ 2π × 0.2236 ≈ 1.404 seconds
2. Maximum Kinetic Energy of the Mass
The maximum kinetic energy (KE) of a mass in simple harmonic motion occurs when the mass passes through the equilibrium position. This can be calculated using the formula:
KE_max = 0.5 * m * v_max²
To find the maximum velocity (v_max), we can use the relationship between amplitude (A) and maximum velocity:
v_max = A * ω
Where:
- A is the amplitude (0.05 m).
- ω is the angular frequency, calculated as ω = √(k/m).
First, let's calculate ω:
ω = √(10 N/m / 0.5 kg) = √(20) ≈ 4.472 rad/s
Now, substituting the values to find v_max:
v_max = 0.05 m * 4.472 rad/s ≈ 0.2236 m/s
Now we can find the maximum kinetic energy:
KE_max = 0.5 * 0.5 kg * (0.2236 m/s)²
Calculating (0.2236 m/s)²:
(0.2236)² ≈ 0.050
Thus, we have:
KE_max ≈ 0.5 * 0.5 * 0.050 ≈ 0.0125 J
Summary of Results
In summary, for the mass of 0.5 kg oscillating with an amplitude of 0.05 m and a spring constant of 10 N/m:
- The period of oscillation is approximately 1.404 seconds.
- The maximum kinetic energy of the mass is approximately 0.0125 Joules.
This analysis illustrates the relationship between mass, spring constant, and the characteristics of harmonic motion, providing a clear understanding of how these elements interact in a simple harmonic oscillator.