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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

The figure shows a mass \(m_{2}=20.0\) g resting on top of a mass \(m_{1}=20.0 \mathrm{~g}\) which is attached to a spring with \(k=10.0 \mathrm{~N} / \mathrm{m}\) The coefficient of static friction between the two masses is 0.600 . The masses are oscillating with simple harmonic motion on a frictionless surface. What is the maximum amplitude the oscillation can have without \(m_{2}\) slipping off \(m_{1} ?\)

A mass \(m=1.00 \mathrm{~kg}\) in a spring-mass system with \(k=\) \(1.00 \mathrm{~N} / \mathrm{m}\) is observed to be moving to the right, past its equi. librium position with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\) at time \(t=0 .\) a) Ignoring all damping, determine the equation of motion. b) Suppose the initial conditions are such that at time \(t=0\), the mass is at \(x=0.500 \mathrm{~m}\) and moving to the right with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\). Determine the new equation of motion. Assume the same spring constant and mass.

A shock absorber that provides critical damping with \(\omega_{\gamma}=72.4 \mathrm{~Hz}\) is compressed by \(6.41 \mathrm{~cm} .\) How far from the equilibrium position is it after \(0.0247 \mathrm{~s} ?\)

Consider two identical oscillators, each with spring constant \(k\) and mass \(m\), in simple harmonic motion. One oscillator is started with initial conditions \(x_{0}\) and \(v_{j}\) the other starts with slightly different conditions, \(x_{0}+\delta x\) and \(v_{0}+\delta v_{1}\) a) Find the difference in the oscillators' positions, \(x_{1}(t)-x_{2}(t)\) for all t. b) This difference is bounded; that is, there exists a constant \(C\) independent of time, for which \(\left|x_{1}(t)-x_{2}(t)\right| \leq C\) holds for all \(t\). Find an expression for \(C\). What is the best bound, that is, the smallest value of \(C\) that works? (Note: An important characteristic of chaotic systems is exponential sensitivity to initial conditions; the difference in position of two such systems with slightly different initial conditions grows exponentially with time. You have just shown that an oscillator in simple harmonic motion is not a chaotic system.)

A mass of \(10.0 \mathrm{~kg}\) is hanging by a steel wire \(1.00 \mathrm{~m}\) long and \(1.00 \mathrm{~mm}\) in diameter. If the mass is pulled down slightly and released, what will be the frequency of the resulting oscillations? Young's modulus for steel is \(2.0 \cdot 10^{11} \mathrm{~N} / \mathrm{m}^{2}\)
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