A particle is moving in a straight line with simple harmonic motion. The period of the motion is (3pi/5)seconds and the amplitude is 0.4metres. Calculate the maximum speed of the particle.
Question: A particle is moving in a straight line with simple harmonic motion. The period of the motion is (3pi/5)seconds and the amplitude is 0.4 metres. Calculate the maximum speed of the particle. Anwer: v2=ω2(a2-x2), v is the linear velocity of the particle, ω is the angular velocity of the particle, a is the amplitude and x is the distance between the position of the particle and the point of centre of oscillation. The maximum speed of the particle will occur when x is 0. Therefore v=ωa by taking the square root of both sides. we know a=o.4 so v=0.4ω. Now we need to find ω.we also know that ω=2pi/(period of the motion)=(2pi)/(3pi/5)=10/3. Now we know the value of ω. Now we can calculate v, v=(0.4ω)=0.4*10/3=4/3 metres per second.I know that I assumed the formula v2=ω2(a2-x2) but I can derive it if needed although it should be proven in the a level text books anyway.
