Question

The amplitude and frequency of an object executing Simple harmonically are 0.01m and 12Hz
respectively. What is the velocity of the object at displacement 0.005m? What is the maximum
velocity of the object?

Answer 1

Answer

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Correct option is

C

0.5

Maximum acceleration

α=ω

2

y

∴0.5=ω

2

×0.02

or ω

2

=

0.02

0.5

=25

So, ω=5

Now, maximum velocity is

v=aω=0.1×5=0.5m/s

Answer 2

The equation of simple harmonic motion can be written in the form x(t) = A sin (ωt + φ),

(where A is the amplitude,

ω is the angular frequency,

and φ is the phase.)

The angular frequency is related to the ordinary frequency f by the formula ω = 2πf.

Now, x(t) is the displacement of object, while its velocity is given by the derivative

v(t)=x'(t)=Aωcos(ωt+φ)=±Aω√(1-〖sin〗^2 (ωt+φ) )=±ω√(A^2-x^2 (t) ).

*The sign here depends on the phase of the motion.

The absolute velocity v at a given displacement x is thus given by v=ω√(A^2-x^2 )=2πf√(A^2-x^2 ).

Substituting here

A = 0.01 m, x = 0.005 m, and f = 12 Hz = 12 s–1, we obtain,

approximately, v = 0.653 m/s.

The maximum velocity vm is reached at x = 0 and is given by

vm = 2πfA = 0.754 m/s.

Answer: v = 0.653 m/s,

vm = 2πfA = 0.754 m/s.