An object executes simple harmonic motion with an amplitude a. (use any variable or symbol stated above as necessary.) (a) at what values of its position does its speed equal half its maximum speed? x = ± (b) at what values of its position does its potential energy equal half the total energy?

(a) x=ASin(ωt+Ф₀)=±(√3)A/2

(b) x=±(√2)A/2

Explanation:

For part (a)

V=AωCos(ωt+Ф₀)⇒±0.5Aω=AωCos(ωt+Ф₀)

Cos(ωt+Ф₀)=±0.5⇒ωt+Ф₀=π/3,2π/3,4π/3,5π/3

x=ASin(ωt+Ф₀)=±(√3)A/2

For part(b)

U=0.5E and U+K=E→K=0.5E

E=K(Max)

(1/2)mv²=(0.5)(1/2)m(Vmax)²

V=±(√2)Vmax/2→ωt+Ф₀=π/4,3π/4,7π/4

x=±(√2)A/2

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  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂