Abody is moving with simple harmonic motion. it's velocity is recorded as being 3.5m/s when it is at 150mm from the mid-position and 2.5m/s when 225mm from mid-position. find : i) it's max amplitude ii) max acceleration iii) the periodic time iv) the frequency of oscillation.

1) A=282.6 mm

2)

3)T=0.42 sec

4)f= 2.24 Hz

Explanation:

Given that

V=3.5 m/s at x=150 mm     ------------1

V=2.5 m/s at x=225 mm   ------------2

Where x measured  from mid position.

We know that velocity in simple harmonic given as

Where A is the amplitude and ω is the natural frequency of simple harmonic motion.

From equation 1 and 2

   ------3

  --------4

Now by dividing equation 3 by 4

So    A=0.2826 m

A=282.6 mm

Now by putting the values of A in the equation 3

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

Time period T

T=0.42 sec

answer:

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