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: v22(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 v22(a2-x2) but I can derive it if needed although it should be proven in the a level text books anyway.

Answered by Further Mathematics tutor

2757 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

using an integrating factor, find the general solution of the differential equation dy/dx +y(tanx)=tan^3(x)sec(x)


A child weighing 50kg is pushed down a 2m long slide (u=0.1), angled at 45 degrees from the horizontal, at 5m/s. At what speed does the child reach the bottom of the slide?


Find the modulus and argument of the complex number 1+2i


Let A, B and C be nxn matrices such that A=BC-CB. Show that the trace of A (denoted Tr(A)) is 0, where the trace of an nxn matrix is defined as the sum of the entries along the leading diagonal.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning