Question

Need help with part b and c

roblem 14.45 Constants I Periodic Table Part A A 210 g block hangs from a spring with spring constant 6.0 N/m. Att0 s the block is 15 cm below the equilibrium point and moving upward with a speed of 134 cm/s What is the block's oscillation frequency? Express your answer to two significant figures and include the appropriate units. f 0.85 Hz Previous Answers Correct Part B What is the block's distance from equilibrium when the speed is 48 cm/s? Express your answer to two significant figures and include the appropriate units. 0.26m Submit Previous Answers Request Answer

Answer 1

B) here when the time t=0,

the total energy is

E=Epotential+E kinetic........1)

E=1/2kx²+1/2mv²=1/2*6N/m*0.15²+1/2*0.21kg*1.34²=0.256038J

when the new speed=0.48m/s

Ek=1/2*0.21kg*0.48²=0.024192J

so now from eqn 1)

E=Ep-Ek

Ep=E-Ek=0.256038-0.024192=0.231846J

so Ep=1/2kx²

0.231846=1/2*6*x²

x²=0.077282

x=0.27m

so the answer is 0.27m or 0.30m

C) we know the formula for amplitude

x(t)=Acos($\omega$t-$\phi$)

A can be found out by

E=Ep

0.256038J=1/2*6*x²

x=$\sqrt{}$0.085346=29.2cm

now from eqn 1 we have

15=29.2cos(0-$\phi$)

cos(-$\phi$)=0.5137

$\phi$=-cos^-1(0.5137)= -1.031rad

so now when t=10s eqn 1 becomes

x(10s)=29.2cm*cos(5.338*10+1.031)=29 cm

so the answer is 29.0 cm or 28.8 cm