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 R2 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?

try 0.47

try 0.047

or 0.0804

To calculate the absolute value of the difference in periods of oscillation before and after the part fell off, we first need to determine the period of oscillation in each case.

The period of oscillation for a simple pendulum can be defined as:

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 pendulum is represented by the clock, and the length of the pendulum (L) changes after the circular part next to the hanging point falls off.

Before the circular part fell off, the length of the pendulum (L1) is equal to the radius of the disk (R) plus the radius of the circular part (R2). So, L1 = R + R2.

After the circular part fell off, the length of the pendulum (L2) is equal to just the radius of the disk (R). So, L2 = R.

To calculate the periods of oscillation, we need to calculate the lengths of the pendulum in each case.

Now, it is not indicated whether the hanging point remains the same after the circular part falls off. If the hanging point remains the same, the length of the pendulum doesn't change, and the periods of oscillation before and after the part fell off would be the same. In this case, the absolute value of the difference in periods would be zero.

However, if the hanging point changes to coincide with the new edge of the clock after the circular part falls off, the length of the pendulum changes, and the periods of oscillation would be different.

In this case, before the circular part fell off, the length of the pendulum (L1) would be equal to the initial distance from the hanging point to the edge of the disk, which is the radius of the disk (R). So, L1 = R.

After the circular part fell off, the length of the pendulum (L2) would be equal to the distance from the new hanging point to the edge of the disk, which is again the radius of the disk (R). So, L2 = R.

The periods T1 and T2 for each case can be calculated using the formula mentioned earlier:

T1 = 2π√(L1/g) = 2π√(R/g)

T2 = 2π√(L2/g) = 2π√(R/g)

The absolute value of the difference in the periods of oscillation would be:

|T1 - T2| = |(2π√(R/g)) - (2π√(R/g))| = 0

Therefore, if the hanging point remains the same, the absolute value of the difference in the periods of oscillation would be zero.