Question

a body of 0.25 KG executes SHM given by y=4sin 0.5πtm . calculate amplitude, angular frequency, maximum velocity and maximum acceleration.​

Answer 1

Amplitude is 4 m

Angular frequency is 0.5π rad/s

Max Velocity is v = Aw = 4*0.5π = 6.28 m/s

Max accelaration is a= Aw² = 0.25*π²*4 = 9.8 m/s²

Answer 2

Answer:

The amplitude,angular frequency,maximum velocity, and maximum acceleration are 4m,0.5π rad/sec,2πm/s and π²m/s² respectively.

Explanation:

The waveform given in the question is of the form,

$y=Asin\omega t$     (1)

Where,

y=displacement of the body in meter

A=amplitude of the body

ω=angular frequency

t=time during which the wave travels

From the question we have,

Mass of the body(m)=0.25 kg

$y=(4sin0.5\pi t )m$    (2)

By comparing equations (1) and (2) we get;

The amplitude of the body is (A)=4 m

The angular frequency of the body is (ω)=0.5π rad/sec

The maximum velocity is given as(v),

$v=A\omega$     (3)

By substituting the values in equation (3) we get;

$v=4\times 0.5\pi$

$v=2\pi m/s$    (4)

The maximum acceleration is given as,

$a=A\omega^2$     (6)

By substituting the values in equation (6) we get;

$a=4\times (0.5\pi)^2$

$a=\pi^2 m/s^2$    (7)

Hence, the amplitude,angular frequency,maximum velocity, and maximum acceleration are 4m,0.5π rad/sec,2πm/s and π²m/s² respectively.

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