Question

1. A 2.0 kg object is attached to a horizontal spring of force constant k = 4.3 kN/m. The spring is stretched 10 cm from equilibrium and released. Find its total energy.

..... J

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2. Find the total energy of a 2.0 kg object oscillating on a horizontal spring with an amplitude of 10 cm and a frequency of 2.9 Hz.
..... J

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3. Find the length of a simple pendulum if its frequency for small amplitudes is 1.00 Hz.
........ cm

Answer 1

1. E = ( 1/ 2) kx ²

E =(1/2)(4.3)(0.1)^2
E = 0.0215 joules

2.

Mass m = 2 kg

Amplitude A = 10 cm = 0.1 m

Frequency f = 2.9 Hz

Angular frequency ? = 2?f

                                = 18.222 rad / s

Total energy of the object E = ( 1/ 2) m? ² A ²

                                            = 3.32 J

3.

Given:
the frequency of simple pendulum f = 1.0 Hz

solution:

           the frequency of a simple pendulum is

                              f =(1/2pi) root g/L   ........ (1)

     therefore the length of the a simple pendulum is

                              L =g/(2pif)^2 ......... (2)   (from equation '1')

where the acceleration due to gravity g = 9.8 m/s^2

                                                         pi=3.14

substitute the given data in equation (2),we get

                         L=(9.8 m/s^2)/[(2)(3.14)(1.0 Hz)]^2

   the length of the a simple pendulum L = 1.56 m = 156 cm