A pendulum located where has a 0.500 kg bob and a 4.00 Hz oscillation frequency.If it is released from a point 0.150 m above the lowest point it reaches, the number of quanta in this oscillation is:
A)2.77 * 1032.
B)1.11 * 1033.
C)4.87 * 10-34.
D)0.184.
E)4.00.

Question

A pendulum located where has a 0.500 kg bob and a 4.00 Hz oscillation frequency.If it is released from a point 0.150 m above the lowest point it reaches, the number of quanta in this oscillation is: 
A)2.77 * 1032. 
B)1.11 * 1033. 
C)4.87 * 10-34. 
D)0.184. 
E)4.00.

Quizplus · Accepted Answer

The number of quanta, or energy packets, in an oscillation can be calculated using the equation:

E = nhf

where E is the energy of each packet, n is the number of packets, h is Planck's constant, and f is the frequency of the oscillation.

Solving for n:

n = E / hf

The energy of each packet can be found using the equation:

E = 1/2 m v^2

where m is the mass of the bob and v is the maximum velocity it reaches during oscillation. The maximum velocity can be found using:

v = Aω

where A is the amplitude of oscillation and ω is the angular frequency, which is equal to 2πf.

Substituting the given values:

ω = 2πf = 2π(4.00 Hz) = 25.13 rad/s

v = (0.150 m)(25.13 rad/s) = 3.77 m/s

E = 1/2 (0.500 kg) (3.77 m/s)^2 = 3.56 J

Substituting into the previous equation:

n = (3.56 J) / [(6.63 x 10^-34 J s) (4.00 Hz)] = 2.77 x 10^32

A Pendulum Located Where Has a 0