Question #308589

An object oscillates with simple harmonic motion along the x-axis, its position varies with time. According to the equation. x(t) = 4 Cos(πt + π/3 ). The time is given in a sec and the angles are in radian. a) Draw a free hand wave diagram. b) Determine the amplitude, frequency, and time period of the motion. c) Calculate the velocity and acceleration of the object at any time ‘t’. d) Using the result of part ‘b’,find velocity and acceleration at t = 5 sec.

1
Expert's answer
2022-03-10T10:18:03-0500

Solution.

x(t)=4cos(πt+π3);x(t)=4cos(\pi t+\dfrac{\pi}{3});

b)x(t)=Acos(wt+ϕ0);b) x(t)=Acos(wt+\phi _0);

A=4m;w=πrad;A=4m; w=\pi rad;

w=2πf    f=w2π;w=2\pi f\implies f=\dfrac{w}{2\pi};

f=π2π=12s1;f=\dfrac{\pi}{2\pi}=\dfrac{1}{2} s^{-1};

T=1f;T=\dfrac{1}{f}; T=10.5=2s;T=\dfrac{1}{0.5}=2s;

c)v(t)=x(t)=4πsin(πt+π3);c) v(t)=x^{'}(t)=-4\pi sin(\pi t+\dfrac{\pi}{3});

a(t)=v(t)=4π2cos(πt+π3);a(t)=v^{'}(t)=-4\pi^2cos(\pi t+\dfrac{\pi}{3});

d)v(5)=43.14sin(5π+π3)=10.87m/s;d) v(5)=-4\sdot 3.14\sdot sin(5\pi+\dfrac{\pi}{3})=10.87 m/s;

a(5)=43.142cos(5t+π3)=19.72m/s2;a(5) =-4\sdot 3.14^2\sdot cos(5t+\dfrac{\pi}{3})=19.72 m/s^2;

Answer: b)f=0.5s1;T=2s;b) f=0.5s^{-1}; T=2s;

c)v(t)=4πsin(πt+π3);a(t)=4π2cos(πt+π3);c)v(t)=-4\pi sin(\pi t+\dfrac{\pi}{3});a(t)=-4\pi^2cos(\pi t+\dfrac{\pi}{3});

d)v(5)=10.87m/s;a(5)=19.72m/s2.d)v(5)=10.87 m/s; a(5)=19.72 m/s^2.


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