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Chapter 14: Problem 69

A grandfather clock uses a pendulum and a weight. The pendulum has a period of \(2.00 \mathrm{~s}\), and the mass of the bob is 250. \(\mathrm{g}\). The weight slowly falls, providing the energy to overcome the damping of the pendulum due to friction. The weight has a mass of \(1.00 \mathrm{~kg}\), and it moves down \(25.0 \mathrm{~cm}\) every day. Find \(Q\) for this clock. Assume that the amplitude of the oscillation of the pendulum is \(10.0^{\circ}\)

Short Answer

Expert verified
Answer: The quality factor (Q) for the given grandfather clock is approximately 3760.

Step by step solution

01

Calculate the initial total energy of the pendulum

We have the mass of the bob (m_bob = 250 g, which is 0.25 kg), and the amplitude of oscillation (10°). We'll first convert the angle to radians: 10° × (π/180) = 10π/180 ≈ 0.174 radians We need to find the length of the pendulum (l) using the period formula T = 2π √(l/g), where g is the gravitational acceleration (9.81 m/s²). We will rearrange the formula to calculate l: l = (T^2 * g) / (4π^2) Given T = 2 seconds, the length of the pendulum can be calculated as: l ≈ ((2^2 * 9.81) / (4π^2)) ≈ 1.005 m The initial total energy E can be calculated as the gravitational potential energy E = m_bob * g * h, where h = l * (1 - cos(amplitude)). E ≈ 0.25 * 9.81 * 1.005 * (1 - cos(0.174)) ≈ 0.034 J (Joules)
02

Calculate the energy lost due to friction each day

The weight of the clock moves downward 25 cm (0.25 m) every day. We can find the gravitational potential energy lost by the weight each day as E(f) = m_weight * g * h_movement, where m_weight = 1.00 kg and h_movement = 0.25 m. E(f) ≈ 1 * 9.81 * 0.25 ≈ 2.453 J (Joules)
03

Calculate the energy lost per cycle

To find the energy lost during each cycle, we need to know how many cycles the pendulum completes in a day. There are 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute, so the total seconds in a day are: Total_seconds = 24 * 60 * 60 = 86,400 s The period of the pendulum is 2.00 s, so the number of cycles completed in a day is: No_of_cycles = Total_seconds / T = 86,400 / 2 = 43,200 cycles Thus, the energy lost per cycle e_lost can be calculated as: e_lost = E(f) / No_of_cycles = 2.453 J / 43,200 ≈ 5.676 × 10^-5 J
04

Calculate the quality factor (Q) for the clock

Now that we have the initial total energy (E) and the energy lost per cycle (e_lost), we can find the quality factor Q using the formula: Q = 2π * (E / e_lost) Q ≈ 2π * (0.034 J / 5.676 × 10^-5 J) ≈ 3760 The quality factor (Q) for this grandfather clock is approximately 3760.

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Most popular questions from this chapter

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