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Nostrana [21]
3 years ago
8

A body is moving with simple harmonic motion. It's velocity is recorded as being 3.5m/s when it is at 150mm from the mid-positio

n 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.
Engineering
1 answer:
natima [27]3 years ago
8 0

Answer:

1) A=282.6 mm

2)a_{max}=60.35\ m/s^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

V=\omega \sqrt{A^2-x^2}

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

From equation 1 and 2

3.5=\omega \sqrt{A^2-0.15^2}    ------3

2.5=\omega \sqrt{A^2-0.225^2}   --------4

Now by dividing equation 3 by 4

\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}

1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}

So    A=0.2826 m

A=282.6 mm

Now by putting the values of A in the equation 3

3.5=\omega \sqrt{A^2-0.15^2}

3.5=\omega \sqrt{0.2826^2-0.15^2}

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

a_{max}=\omega ^2A

a_{max}=14.61 ^2\times 0.2826\ m/s^2

a_{max}=60.35\ m/s^2

Time period T

T=\dfrac{2\pi}{\omega}

T=\dfrac{2\pi}{14.609}

T=0.42 sec

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