Question

The position of a mass (350 g) attached to an oscillating spring is given by:

x = 22.5 cm cos((7.84 rad/s) t)

1. Find total energy of the mass.

2. Determine the potential energy when the mass is located 5.3 cm from equilibrium.

3. What is the velocity of the mass at the location in part B?

4. Find the location of the mass when the velocity is one-third of its maximum value.

Answer 1

(0.) 6tal ehevzy of Simple hormonic mohion is deined as musA2 w amguln frequeny A Amplitude mzmas GIVen equatn 0225m Cas7.84Yadls)t X A m-0.350, theu (0350 74 radlsC0-225m -

E 0.5445 J bPotential cher3y- is defined ab 2. C03507-84xad (sC0.053m theu U U= 0-0302丁 (c) 0-225m Cs7-34t) Xニ

Wheu xO. 053 m i hen 0-225m Ces (7.89 t) O053M O.053m O-169947ad 7-84 Yad.js O.225m 7-By Sin (7.84t) V= _ -0.225x7.84 V -76 ms Sin (7.84 t) wheu t= 0-699 4E ea Now,

V -17 mbn (7.34 x0-169 Tad (polEE11jmh9L;( - =A (A)V m4x Vma Vmar

0.34966 rad 7.81 TaAls 6.043324 Sea CN オ=0-043324 Se2 0.22sm C (7-84 x 0.043429%ad.) X 21.2 Cm 6r 43