Answer:
The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%
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;
Therefore, the percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%
An impulsive force is a force that is acting only during a short time, I mean, for an instant. Impulse is a physics magnitude define by the product of the impulsive force and the time that it was acting.
Is there any mistake in my English? Please, let me know.
For this problem, we use the decay equation which is expressed as:
An = (Ao)e^-kt
where An is the amount left in t time, Ao is the initial amount and k is a constant.
We first calculate the value of k as follows:
An = (Ao)e^-kt
0.5 = e^-k(8)
k = 0.087
We calculate the time as follows:
An = (Ao)e^-kt
15.625 = 250e^-0.087t
t = 31.87 days