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Chapter 11: Problem 46

\(\bullet\) You've made a simple pendulum with a length of 1.55 \(\mathrm{m}\) , and you also have a (very light) spring with force constant 2.45 \(\mathrm{N} / \mathrm{m} .\) What mass should you add to the spring so that its period will be the same as that of your pendulum?

Short Answer

Expert verified
The mass should be approximately 0.3878 kg.

Step by step solution

01

Determine the Period of the Pendulum

The period of a simple pendulum is given by the formula \( T_p = 2\pi \sqrt{\frac{L}{g}} \), where \( L = 1.55 \text{ m} \) and \( g = 9.8 \text{ m/s}^2 \). Substitute these values to find \( T_p \).
02

Calculate the Pendulum's Period

Substitute the given values into the formula: \[ T_p = 2\pi \sqrt{\frac{1.55}{9.8}} \] Now compute to find \( T_p \).
03

Determine the Period of the Spring-Mass System

The period of a mass-spring system is given by \( T_s = 2\pi \sqrt{\frac{m}{k}} \), where \( k = 2.45 \text{ N/m} \). We want \( T_s = T_p \).
04

Solve for the Mass Required

Set \( T_s = T_p \) and use the equation \( T_p = 2\pi \sqrt{\frac{m}{2.45}} \). Solve for \( m \) by equating and equating it to \( T_p = 2\pi \sqrt{\frac{1.55}{9.8}} \).
05

Calculate the Required Mass

Square both sides of the equation to remove the square roots and solve for \( m \):\[ \frac{1.55}{9.8} = \frac{m}{2.45} \] Cross-multiply and solve:\[ m = \frac{1.55 imes 2.45}{9.8} \]Evaluate to find the value of \( m \).
06

Final Calculation

By performing the calculations, we find \( m \approx 0.3878 \text{ kg} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pendulum Period
The period of a pendulum is the time it takes for one complete back-and-forth swing. To find this, you can use the formula: \[ T_p = 2 \pi \sqrt{\frac{L}{g}} \]Here, \(T_p\) is the period, \(L\) is the pendulum length, and \(g\) is the acceleration due to gravity, typically \(9.8 \text{ m/s}^2\).
This formula comes from the physics of motion and is a great way to predict how long a pendulum will take to complete a cycle.
  • Longer pendulums take more time per swing.
  • Shorter pendulums swing faster.
Understanding this concept is key for comparing period times between different systems.
Spring Force Constant
The spring force constant, often denoted as \(k\), describes how stiff a spring is. It tells us how much force is needed to stretch or compress the spring by a unit of length. The formula relating force \(F\) to the spring constant \(k\) and displacement \(x\) is:\[ F = kx \]
  • A high \(k\) value means a stiffer spring, requiring more force to compress or extend it.
  • A low \(k\) value means a looser spring, needing less force for displacement.
For our problem, knowing the spring constant (here \(2.45 \text{ N/m}\)) helps determine how much mass will result in a matching swing period when compared to a pendulum.
Mass-Spring System
A mass-spring system consists of a mass attached to a spring. It oscillates due to the force exerted by the spring. The frequency and period of this system are influenced by the mass and the spring's force constant.The period \(T_s\) of such a system is calculated using:\[ T_s = 2\pi \sqrt{\frac{m}{k}} \]Here \(m\) is the mass, and \(k\) is the spring constant.
  • Increasing the mass \(m\) lengthens the period.
  • Increasing the spring constant \(k\) shortens the period.
In our exercise, we need to find the mass that gives a period matching a pendulum, using its corresponding spring force constant.
Period Matching
Period matching is the process of making two different oscillating systems have identical periods. In physics problems like these, it's a fun challenge to create conditions where a pendulum and a mass-spring system oscillate at the same rate.To achieve this, the periods of both systems, given by their formulas, need to be set equal. In our exercise:\[ T_p = T_s \]This leads us to equate the pendulum's period formula with that of the mass-spring system:\[ 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{m}{k}} \]By solving this equation, we can find the necessary mass \(m\) to match the period of the pendulum, resulting in synchronized motions.
Physics Problem-Solving
Solving physics problems often involves identifying the correct equations and understanding the relationships between different variables. Here we deal with oscillations.To solve this exercise:1. Calculate the period of the pendulum using the given length.2. Use the spring constant and find the needed mass for the spring system.3. Set the two periods equal for period matching.4. Solve for the unknown variable, in this case, the mass \(m\).
  • Break the problem into manageable steps.
  • Use clear, logical reasoning.
With practice, these problem-solving steps become almost automatic, making tackling similar physics puzzles easier.

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Most popular questions from this chapter

\(\bullet\) In the Challenger Deep of the Marianas Trench, the depth of seawater is 10.9 \(\mathrm{km}\) and the pressure is \(1.16 \times 10^{8}\) Pa (about 1150 atmospheres). (a) If a cubic meter of water is taken to this depth from the surface (where the normal atmospheric pressure is about \(1.0 \times 10^{3} \mathrm{Pa}\) , what is the change in its volume? Assume that the bulk modulus for seawater is the same as for freshwater \(\left(2.2 \times 10^{9} \mathrm{Pa}\right) .\) (b) At the surface, seawater has a density of \(1.03 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} .\) What is the density of sea-water at the depth of the Challenger Deep?

\(\bullet\) A steel wire has the following properties: $$\begin{array}{l}{\text { Length }=5.00 \mathrm{m}} \\ {\text { Cross- sectional area }=0.040 \mathrm{cm}^{2}} \\ {\text { Young's modulus = } 2.0 \times 10^{11} \mathrm{Pa}} \\ {\text { Shear modulus = } 0.84 \times 10^{11} \mathrm{Pa}} \\ {\text { Proportional limit }=3.60 \times 10^{8} \mathrm{Pa}} \\\ {\text { Breaking stress }=11.0 \times 10^{8} \mathrm{Pa}}\end{array}$$ The wire is fastened at its upper end and hangs vertically. (a) How great a weight can be hung from the wire without exceeding the proportional limit? (b) How much does the wire stretch under this load? (c) What is the maximum weight that can be supported?

\(\bullet\) \(\bullet\) Stress on the shinbone. The compressive strength of our bones is important in everyday life. Young's modulus for bone is approximately 14 GPa. Bone can take only about a 1.0\(\%\) change in its length before fracturing. If Hooke's law were to hold up to fracture: (a) What is the maximum force that can be applied to a bone whose minimum cross-sectional area is 3.0 \(\mathrm{cm}^{2} .\) . This is approximately the cross-sectional area of a tibia, or shinbone, at its narrowest point.) (b) Estimate the maximum height from which a 70 \(\mathrm{kg}\) man can jump and not fracture the tibia. Take the time between when he first touches the floor and when he has stopped to be \(0.030 \mathrm{s},\) and assume that the stress is distributed equally between his legs.

\(\bullet\) Biceps muscle. A relaxed biceps muscle requires a force of 25.0 \(\mathrm{N}\) for an elongation of \(3.0 \mathrm{cm} ;\) under maximum tension, the same muscle requires a force of 500 \(\mathrm{N}\) for the same elongation. Find Young's modulus for the muscle tissue under each of these conditions if the muscle can be modeled as a uniform cylinder with an initial length of 0.200 \(\mathrm{m}\) and a cross-sectional area of 50.0 \(\mathrm{cm}^{2}.\)

\(\bullet\) (a) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern having the same frequency of the note that is sung. If someone sings the note \(B\) flat that has a frequency of 466 \(\mathrm{Hz}\) , how much time does it take the person's vocal cords to vibrate through one complete cycle, and what is the angular frequency of the cords? (b) Hearing. When sound waves strike the eardrum, this membrane vibrates with the same frequency as the sound. The highest pitch that typical humans can hear has a period of 50.0\(\mu \mathrm{s} .\) What are the frequency and angular frequency of the vibrating eardrum for this sound? (c) Vision. When light having vibrations with angular frequency ranging from \(2.7 \times 10^{15} \mathrm{rad} / \mathrm{s}\) to \(4.7 \times 10^{15} \mathrm{rad} / \mathrm{s}\) strikes the retina of the eye, it stimulates the receptor cells there and is perceived as visible light. What are the limits of the period and frequency of this light? (d) Ultrasound. High-frequency sound waves (ultrasound) are used to probe the interior of the body, much as x rays do. To detect a small objects such as tumors, a frequency of around 5.0 \(\mathrm{MHz}\) is used. What are the period and angular frequency of the molecular vibrations caused by this pulse of sound?
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