Answer:
The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%
Step-by-step explanation:
Mechanical energy (Potential energy, PE) of the oscillator is calculated as;
PE = ¹/₂KA²
During the first oscillation;
PE₁ = ¹/₂KA₁²
During the second oscillation;
A₂ = A₁ - 0.0342A₁ = 0.9658A₁
PE₂ = ¹/₂KA₂²
PE₂ = ¹/₂K (0.9658A₁)²
PE₂ = (0.9658²)¹/₂KA₁²
PE₂ = (0.9328)¹/₂KA₁²
PE₂ = 0.9328PE₁
Percentage of the mechanical energy of the oscillator lost in each cycle;
![Change \ in \ percent= (PE_2 - PE_1)/(PE_1) \\\\Change \ in \ percent= (0.9328PE_1 -PE_1)/(PE_1) *100\\\\Change \ in \ percent= (-0.0672PE_1)/(PE_1)*100 \\\\Change \ in \ percent= -0.0672*100\\\\Change \ in \ percent= -6.72 \ \% \\\\Loss \ in \ percent= 6.72 \ \%](https://img.qammunity.org/2021/formulas/physics/college/e3srl5t2i5qs9kzz1sii0xtpwzkevp75p9.png)
Therefore, the percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%