A body of mass 10g oscillates with Simple Harmonic Motion of amplitude 4cm and time period 0.50s.
Calculate:
a) The maximum acceleration.
b) The maximum kinetic energy.
To answer these questions, we will make use of the equations related to simple harmonic motion. Let's break down the problem step by step:
a) The maximum acceleration can be determined using the formula for maximum acceleration in simple harmonic motion, given by:
amax = ω^2 * A
Where:
amax is the maximum acceleration
ω (omega) is the angular frequency of the motion
A is the amplitude
To find ω, we can use the formula for angular frequency:
ω = 2π / T
Where:
T is the time period of the motion
Given that the amplitude A is 4 cm (or 0.04 m) and the time period T is 0.50 s, we can calculate the answers:
First, let's find ω:
ω = 2π / T
ω = 2π / 0.50
ω ≈ 12.57 rad/s
Next, we can find amax:
amax = ω^2 * A
amax = (12.57)^2 * 0.04
amax ≈ 2.51 m/s^2
Therefore, the maximum acceleration is approximately 2.51 m/s^2.
b) To find the maximum kinetic energy (KEmax), we can use the equation for kinetic energy in simple harmonic motion:
KEmax = 1/2 * m * (ω * A)^2
Where:
KEmax is the maximum kinetic energy
m is the mass of the body
Given that the mass of the body is 10 g (or 0.01 kg), we can calculate the answer:
KEmax = 1/2 * m * (ω * A)^2
KEmax = 1/2 * 0.01 * (12.57 * 0.04)^2
KEmax ≈ 0.008 J
Therefore, the maximum kinetic energy is approximately 0.008 Joules.