Question

A 750-gram block is attached to a spring as shown in the following diagram. The system is placed on a horizontal surface. The block is released at a distance of 0.15 m from the equilibrium position at Xo. It oscillates back and forth with a frequency of 0.25 Hz. Assume that the surface is frictionless. The oscillation is an SHM. (a) Find the spring constant. (b) Find the elastic P.E. in the system when the block is at the maximum displacement. (c) Find the total mechanical energy when the block is 0.10 m from the equilibrium position (x = xo + 0.10 m).

Answer 1

when block is released 0.15 m distance from the equilibrium bosition n. Then it begain do do SHM. and eqnis, Masinut tak origin at kung kom Since frequency fi is defined as (9). Hence Spouing constant is, 0.05 - J Verso & K = 0.05x07 r0.750 → K = 1.8496 N/m 5) K1-85 Nm 2. when it is at maximum dispocement ce %=0.Sm. Henel elastic potentiel energy P.E. = 2K22 => REFx1.85 (0.15) -PE=0.020796 T TP.E. = 0.091 Jowel the total mechanical energy CK.EXP.E.) is Constant at ang point of simple hasmonic motion Hence tobal mechanical energy X Rotolom is, P. E. I max*(K. E.) = 0.021 Joule = E=0.021 Jowe

you can also find the mechanical energy by equation of 3.4.m. r- u.sinut 6-a.w cosut v-awut (%) ezwa na - voda22 Hence. Mechanical energy CK E+P. E at w 24 +0.1om. & E= fm u 2 + 2kx? > E-5207508 (004 (224) +LxKx? SE 10.75 min) (615326_149+*+/-85(6-1032 E=0.og) Joule