A clock is made out of a disk of radius R=10 cm which is hung by a point on its edge and oscillates. All of a sudden, a circular part right next to the hanging point of radius R/2 falls off, but the clock continues oscillating. What is the absolute value of the difference in s between the periods of oscillation before and after the part fell off?
1.193
Please verify
no its wrong!
0.47
Please verify
0.080, i guess?
0.0804, i think?
To find the absolute value of the difference in the periods of oscillation before and after the part fell off, we need to understand how the period of oscillation is related to the radius of the clock.
The period of oscillation of a simple pendulum is given by the formula:
T = 2π√(L/g)
Where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
In this case, the clock is hanging from a point on its edge, so the length of the pendulum is equal to the radius of the clock, which is 10 cm.
Before the part fell off, the radius of the clock was 10 cm, so the length of the pendulum was also 10 cm.
T1 = 2π√(10/g)
After the part fell off, the radius of the clock is reduced to half, which means the new length of the pendulum is 5 cm.
T2 = 2π√(5/g)
To find the absolute value of the difference in s between the periods, we need to subtract T2 from T1:
|T1 - T2| = |2π√(10/g) - 2π√(5/g)|
Since both terms have a common factor of 2π, we can factor that out:
|T1 - T2| = 2π|√(10/g) - √(5/g)|
Now, we need the value of g, which is the acceleration due to gravity. On Earth, g is approximately 9.8 m/s².
Converting the lengths from cm to m:
|T1 - T2| = 2π|√(0.1/(9.8)) - √(0.05/(9.8))|
Evaluating the expression:
|T1 - T2| = 2π|√(0.01/9.8) - √(0.005/9.8)|
|T1 - T2| ≈ 2π|0.0316 - 0.0224|
|T1 - T2| ≈ 2π|0.0092|
|T1 - T2| ≈ 0.058 s
Therefore, the absolute value of the difference in s between the periods of oscillation before and after the part fell off is approximately 0.058 seconds.