A simple harmonic oscillator executes motion whose amplitude is 0.20m and it completes 60 oscillations in 2 mints.
1. Calculate its time period and angular frequency.
2. If the initial phase is 45°, write expressions for instantaneous displacement, velocity and acceleration.
3. Also calculate the maximum values of velocity and acceleration of the oscillator.

1

Answers

2015-04-02T03:58:24+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Simple harmonic motion:

       A = amplitude = 0.20 m   
       60 oscillations in 2 minutes = 120 sec.
       So frequency = 1/2 oscillation in 1 sec.  => f = 1/2 Hz
           time period T = 1/f = 2 seconds.
           angular frequency = ω = 2 π f = 2 π / T = π radians
==================
2.    let the displacement from the mean position during the SHM be:
                 x (t) = A Cos (ω t + Ф)
                        = 0.20 Cos ( π t + π/4)  meters
                        =  020 Cos π (t + 0.25)  meters

         velocity v(t) = d x(t) / dt  = - 0.20 π Sin π(t + 0.25)

       acceleration = a(t) = d v(t) / dt = - 0.20 π² Cos π(t + 0.25)
                     or,    a (t) = - π² x(t)
============================
 3.      maximum value of velocity = v(t = 0.25 sec or 1.25 sec)
               = 0.20 π  m/sec

       maximum value of acceleration = a(t = 0.75 sec or 1.75 sec)
               = 0.20 π²  m/sec²

4 3 4