Answer to Question #96348 in Mechanics | Relativity for Fahadhos

Question #96348
a body of mass 50 gram is attached to a spring. the motion of the spring will have the amplitude 12 centimetre and the time period 1.7 second .find the the maximum velocity, maximum acceleration, velocity at time 5 sec and acceleration at distance 0.06m
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Expert's answer
2019-10-14T09:17:11-0400

The maximum velocity:


V=Aω=2πAT=2π(0.12)1.7=0.44msV=A\omega=\frac{2\pi A}{T}=\frac{2\pi (0.12)}{1.7}=0.44\frac{m}{s}

The maximum acceleration:


V=Aω2=A(2πT)2=0.12(2π1.7)2=1.6ms2V=A\omega^2=A\left(\frac{2\pi }{T}\right)^2=0.12\left(\frac{2\pi }{1.7}\right)^2=1.6\frac{m}{s^2}

The velocity at time 5 sec:

v(5)=0.44sin(2π(5)1.7)=0.16msv(5)=0.44\sin{\left(\frac{2\pi (5)}{1.7}\right)}=-0.16\frac{m}{s}

The acceleration at distance 0.06 m:

a(5)=0.06(2π1.7)2=0.82ms2a(5)=-0.06\left(\frac{2\pi }{1.7}\right)^2=-0.82\frac{m}{s^2}


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