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Chapter 2: Problem 4

If a mass \(m\) is attached to a given spring, its period of oscillation is \(T\). If two such springs are connected end to end, and the same mass \(\mathrm{m}\) is attached, find the new period \(T^{\prime}\), in terms of the old period \(T\).

Short Answer

Expert verified
Answer: The new period of oscillation, T', is equal to the old period, T, divided by the square root of 2: \(T' = \frac{T}{\sqrt{2}}\).

Step by step solution

01

Formula for the period of oscillation for a mass-spring system

We can use the formula for the period of oscillation for a mass-spring system, which is given by: \(T = 2\pi\sqrt{\frac{m}{k}}\) where: - T is the period of oscillation - m is the mass attached to the spring - k is the spring constant
02

Determine the spring constants of the two springs connected end-to-end

When two springs are connected end-to-end, their spring constants are connected in series. The equivalent spring constant, k', can be given by the formula: \(\frac{1}{k'} = \frac{1}{k_1} + \frac{1}{k_2}\) In our case, since both springs are identical, we can replace both \(k_1\) and \(k_2\) with k: \(\frac{1}{k'} = \frac{1}{k} + \frac{1}{k}\) Solve for k': \(k' = \frac{k}{2}\)
03

Find the new period of oscillation with the equivalent spring constant

Now that we have found the equivalent spring constant, k', for the two springs connected end-to-end, we can use the formula for the period of oscillation to find the new period, T': \(T' = 2\pi\sqrt{\frac{m}{k'}}\) Substitute the expression for k' from Step 2: \(T' = 2\pi\sqrt{\frac{m}{\frac{k}{2}}}\)
04

Express the new period in terms of the old period

Now we need to express T' in terms of T. To do this, we can use the original period equation, \(T = 2\pi\sqrt{\frac{m}{k}}\). Solve for k: \(k = \frac{4\pi^2m}{T^2}\) Now substitute this expression for k into the equation for T' from Step 3: \(T' = 2\pi\sqrt{\frac{m}{\frac{4\pi^2m}{2T^2}}}\) Now, simplify the expression: \(T' = 2\pi\sqrt{\frac{2T^2}{4\pi^2}}\) \(T' = 2\pi\frac{T}{\sqrt{4}\pi}\) \(T' = \frac{T}{\sqrt{2}}\) So, the new period of oscillation, T', is equal to the old period, T, divided by the square root of 2.

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Key Concepts

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

Mass-Spring System
A mass-spring system is a fundamental concept in physics where a mass is attached to a spring, causing it to oscillate back and forth. This system is often used to study harmonic motion.
The motion is characterized by its period, which is the time it takes for one complete cycle of oscillation.
In this system:
  • Mass (m): This is the object that is attached to the spring. Its mass affects the period of oscillation, with a larger mass generally resulting in a longer period.
  • Spring Constant (k): This measures the stiffness of the spring. A higher spring constant means a stiffer spring, which can lead to a shorter period of oscillation.
Spring Constant
The spring constant, often represented by the letter \( k \), is a measure of a spring's stiffness.
In the context of a mass-spring system, it is crucial for determining the period of oscillation.
For a spring constant:
  • Units: It is measured in Newtons per meter (N/m).
  • Effect on Oscillation: A larger \( k \) results in a faster oscillation, while a smaller \( k \) means slower oscillation.
Knowing \( k \) helps us understand how external forces will affect the system's motion and allows us to predict the system's behavior under various conditions.
Springs in Series
When two springs are connected in series, their combined effect creates a new equivalent spring constant. This equivalent spring constant is always less than the smallest spring constant in the series.
To find the equivalent spring constant \( k' \) for two identical springs connected in series:
  • Formula: \( \frac{1}{k'} = \frac{1}{k} + \frac{1}{k} \).
  • Simplification: This leads to \( k' = \frac{k}{2} \), meaning the system acts as if the springs are less stiff than a single spring.
This behavior is essential in understanding how connecting springs affects the dynamics of a mass-spring system.
Period of Oscillation Formula
The period of oscillation is a crucial concept in a mass-spring system and is determined by both the mass and the spring constant.
The formula for the period \( T \) is:\[ T = 2\pi\sqrt{\frac{m}{k}} \]

Finding the New Period

When two springs are in series, the equivalent spring constant \( k' \) changes. The new period \( T' \) can be derived using:
  • Formula: \( T' = 2\pi\sqrt{\frac{m}{k'}} \).
  • In Terms of Old Period: Substituting the equivalent constant gives \( T' = \frac{T}{\sqrt{2}} \).
This shows how the system's dynamics change with different setups, providing insights into harmonic motion.

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Most popular questions from this chapter

Two blocks, of mass \(m\) and \(2 m\), are connected by a massless string and slide down an inclined plane at angle \(\theta\). The coefficient of kinetic friction between the lighter block and the plane is \(\mu\), and that between the heavier block and the plane is \(2 \mu\). The lighter block leads. (a) Find the magnitude of the acceleration of the blocks. (b) Find the tension in the taut string.

A wooden bar of uniform density but varying thickness hangs suspended on two strings of negligible mass. The strings make angles \(\theta_{1}\) and \(\theta_{2}\) with the horizontal, as shown. The bar has total mass \(m\) and length \(L\). Find the distance \(x\) between the center of mass of the bar and its (thickest) end.

When you cook rice, some of the dry grains always stick to the measuring cup. A common way to get them out is to turn the measuring cup upside-down and hit the bottom (now on top) with your hand so that the grains come off \([32]\). (a) Explain why static friction is irrelevant here. (b) Explain why gravity is negligible. (c) Explain why hitting the cup works, and why its success depends on hitting the cup hard enough.

Round fruits like oranges and mandarins are typically stacked in alternating rows, as shown in figure \(2.9 .\) Suppose you have a crate with a square base that is exactly five oranges wide. You stack 25 oranges in the crate, then put another 16 on top in the holes, and then add a second layer of 25 oranges, held in place by the sides of the crate. Find the total force on the sides of the crate in this configuration. Assume all oranges are spheres with a diameter of \(8.0 \mathrm{~cm}\) and a mass of \(250 \mathrm{~g}\).

A uniform stick of mass \(M\) and length \(L=1.00 \mathrm{~m}\) has a weight of mass \(m\) hanging from one end. The stick and the weight hang in balance on a force scale at a point \(x=20.0 \mathrm{~cm}\) from the end of the stick. The measured force equals \(3.00 \mathrm{~N}\). Find both the mass \(M\) of the stick and \(m\) of the weight.
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