Problem. A mass of 0.25 kg moves executes simple harmonic oscillations. Its disp

Question

Problem. A mass of 0.25 kg moves executes simple harmonic oscillations. Its displacement as function of time is given by x(t)=0.05sin(125t) with all quantities given in SI units

1) The velocity of the mass has a maximum magnitude at which of the following times (in s)?
A) 0 B) 0.125 C)0.187 D)0.206 E)0.274

2)The frequency of the oscillator is ____Hz.
A) 0.05 B) 20 C)126 D)395 E)440

3)When the displacement of the mass is maximum its acceleration has a magnitude ___m/s^2.
A) 0 B) 0.05 C)6.3 D)794 E)2420

4)When the magnitude of the displacement of the mass is minimum, its velocity has a magnitude _______ m/s.
A) 0 B) 0.05 C)6.3 D)794 E)2420

Explanation / Answer

mass m= 0.25 kg
displacement as function of time x(t)=0.05sin(125t) compare this with x(t) = A sin t , we get amplitude i.e, maximum displacment A = 0.05 m angular frequency = 125 rad / s period of time T = 2 / = 0.0502 s velocity function v(t) = dx(t) / dt                                = 0.05 *cos(125 t) * 125                                = 6.25 cos 125t maximum velocity = 6.25 m / s accleration fuction a(t) = dv(t) / dt                                    = 6.25 [-sin125 t ] * 125                                    = - 781.25 sin 125 t maximum accleration = 781.25 m / s^ 2 1) The velocity of the mass has a maximum magnitude at time T / 4 So, answer is = T / 4 = 0.0502 / 4 = 0.125 s option ( B ) is correct (2).The frequency of the oscillator is f = / 2 = 19.89 Hz ~ 20 Hz option(B) is correct
3)When the displacement of the mass is maximum its acceleration = maximum accleration                                                                                                    = 781.25 m /s^2 ~ 794 m / s^ 2 4)When the magnitude of the displacement of the mass is minimum, its velocity has a magnitude is maximum So, answer is = 6.25 m / s

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