Answer to Question #96348 in Mechanics | Relativity for Fahadhos

Question

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
Please answer early as possible

Answer 1

The maximum velocity:

$$V=A\omega=\frac{2\pi A}{T}=\frac{2\pi (0.12)}{1.7}=0.44\frac{m}{s}$$

The maximum acceleration:

$$V=A\omega^2=A\left(\frac{2\pi }{T}\right)^2=0.12\left(\frac{2\pi }{1.7}\right)^2=1.6\frac{m}{s^2}$$

The velocity at time 5 sec:

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

The acceleration at distance 0.06 m:

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