Question

A 0.20 kg mass at the end of a spring oscillates 2.9 times per second with an amplitude of 0.14 m .

Part A Determine the speed when it passes the equilibrium point. Express your answer to two significant figures and include the appropriate units. vmax = 2.55 ms SubmitMy AnswersGive Up Correct Part B Determine the speed when it is 0.12 m from equilibrium. Express your answer to two significant figures and include the appropriate units. v = SubmitMy AnswersGive Up Part C Determine the total energy of the system. Express your answer to two significant figures and include the appropriate units. Etotal =

Answer 1

Given Data

mass m = 0.2 kg

Frequency (f) = 2.9 Hz

Amplitude (A) = 0.14 m

Solution :-

a)

Now formula for calculating the value of the max speed atthe equilibrium point is

          v_max = Aω

                   = A(2πf)

          v_max = 0.14(2π*2.9)

          v_max = 2.55 m/s

b)

Given that x = 0.12 m

Now formula for calculating the value of the speed atthis point (x = 0.12 m) is

        

          v = (2π*f) sqrt(A² - x²)

          v = (2π*2.9) sqrt(0.14² - 0.12²)

          v = 1.31 m/s

c)

Formula for calculating the value of the total energy of tehsystem is

E total = (1/2)mA²ω²  

E total = 0.5*m*A²*4π²f²

E total = 0.5*0.20 *0.14²*4π²2.9²

E total = 0.651 J