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2022

Physics

POST UTME

A body performing simple harmonic motion has a frequency of 10Hz and an amplitude of 50Hz and an amplitude of 5cm. Calculate its velocity at a distance of 4cm from the centre of oscillation. [π\pi = 3.14].

A.

4.02m/s

B.

3.14m/s

C.

2.15m/s

D.

1.88m/s

Correct Answer:

1.88m/s

Explanation


using,v = ωA2x2ω \sqrt {{ A^2 } ​ − {x^2}}

where A = 5cm, x =4cm

ω=2πf\omega = {2\pi f} ; v=2πfA2x2v = {2\pi f} \sqrt {{A^2 - x^2}}

v=2v = 2 x 3.143.14 x105242\sqrt {{5^2} - {4^2}}

v=v = 22 x 3.143.14 x 10251610 \sqrt {{25} - {16}}

v=v =  1.88cm/s

in m/s we divide by 100

v=v =  1.88m/s


using,v = ωA2x2ω \sqrt {{ A^2 } ​ − {x^2}}

where A = 5cm, x =4cm

ω=2πf\omega = {2\pi f} ; v=2πfA2x2v = {2\pi f} \sqrt {{A^2 - x^2}}

v=2v = 2 x 3.143.14 x105242\sqrt {{5^2} - {4^2}}

v=v = 22 x 3.143.14 x 10251610 \sqrt {{25} - {16}}

v=v =  1.88cm/s

in m/s we divide by 100

v=v =  1.88m/s

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