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Chapter 15: Problem 52

A mass \(M\), attached to a spring, oscillates with a period of 2 s. If the mass is increased by \(4 \mathrm{~kg}\), the time period increases by \(1 \mathrm{~s}\). Assuming that Hooke's law is obeyed, the initial mass \(M\) was (a) \(3.2 \mathrm{~kg}\) (b) \(1 \mathrm{~kg}\) (c) \(2 \mathrm{~kg}\) (d) \(8 \mathrm{~kg}\)

Short Answer

Expert verified
The initial mass \( M \) was 3.2 kg (Option a).

Step by step solution

01

Understand the Formula for Period of a Spring-Mass System

The formula for the period \( T \) of a mass \( M \) attached to a spring is given by \( T = 2\pi\sqrt{\frac{M}{k}} \), where \( k \) is the spring constant. Here, the initial period \( T_1 \) is 2 s.
02

Use Given Data for First Scenario

From the problem, we know that initially the period \( T_1 = 2 \) s. Hence, we can write:\[ T_1 = 2\pi\sqrt{\frac{M}{k}} = 2 \]
03

Square the Initial Condition Equation

Square both sides of \( T_1 = 2 = 2\pi\sqrt{\frac{M}{k}} \):\[ 4 = \frac{4\pi^2 M}{k} \]Thus, \( M = \frac{k}{\pi^2} \) (Equation 1)
04

Use Data When Mass is Increased by 4 kg

When the mass is increased by 4 kg, the new period \( T_2 = 3 \) s. Hence, set up the equation:\[ T_2 = 2\pi\sqrt{\frac{M+4}{k}} = 3 \]
05

Square the New Condition Equation

Square both sides of the new period equation:\[ 9 = \frac{4\pi^2(M+4)}{k} \]Rearrange to find an expression for \( M \):\[ M+4 = \frac{9k}{4\pi^2} \] (Equation 2)
06

Solve Simultaneous Equations

Solve Equation 1 and Equation 2:- From Equation 1: \( M = \frac{k}{\pi^2} \)- Substitute into Equation 2: \( \frac{k}{\pi^2} + 4 = \frac{9k}{4\pi^2} \)Multiply through by \( 4\pi^2 \) to clear the fractions:\[ 4k + 16\pi^2 = 9k \]- Solve for \( k \): \( 5k = 16\pi^2 \), hence \( k = \frac{16\pi^2}{5} \)- Substitute \( k \) back to find \( M \):\[ M = \frac{\frac{16\pi^2}{5}}{\pi^2} = \frac{16}{5} = 3.2 \]
07

Verify with Options Given

Option (a) corresponds to the mass of 3.2 kg, which matches our calculated value for the initial mass \( M \). Hence, option (a) is correct.

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Key Concepts

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

Spring Constant
The spring constant, represented by the symbol \( k \), is a measure of the stiffness of a spring. It describes how much force is needed to compress or extend a spring by a unit length.
  • Mathematically, it's expressed in units of force per length, typically Newtons per meter (N/m).
  • The larger the spring constant, the stiffer the spring, meaning more force is required to change its length.
  • Hooke's Law, which states \( F = kx \), helps in understanding the relationship between the force \( F \) applied to a spring and the displacement \( x \) it causes.
In our exercise, the spring constant \( k \) is crucial because it directly affects the period of oscillation of the mass-spring system. The period \( T \) is related to both the mass and the spring constant, as shown in the formula \( T = 2\pi\sqrt{\frac{M}{k}} \). Understanding \( k \) allows us to determine how mass changes influence oscillation speed.
Period of Oscillation
The period of oscillation, \( T \), is the time a mass-spring system takes to complete one full cycle of motion. It depends on two primary factors: the mass \( M \) attached to the spring and the spring constant \( k \).
  • It is calculated using the formula \( T = 2\pi\sqrt{\frac{M}{k}} \).
  • A larger mass results in a longer period, meaning the system takes more time to complete its oscillation due to increased inertia.
  • On the other hand, a stiffer spring (larger \( k \)) leads to a shorter period, making oscillations faster.
In the given exercise, the initial period of the mass-spring system is given as 2 seconds, and it changes to 3 seconds when more mass is added. This information is crucial for solving problems related to oscillating systems, as it reveals the dynamics of how periods adjust with varying masses and spring properties.
Mass-Spring System
The mass-spring system is a classic model in physics used to study oscillatory motion. It consists mainly of a mass that causes a spring to stretch or compress as it moves.
  • When a mass \( M \) is attached to a spring, it creates a simple harmonic oscillator.
  • The system's behavior is governed by Hooke's Law, which links the spring's force to the displacement it caused.
  • The energy within the system constantly shifts between potential energy stored in the spring and the kinetic energy of the moving mass.
Our exercise highlights a mass-spring system undergoing periodic oscillations, showcasing how changes in mass affect the system's dynamics. By understanding this system, students can grasp fundamental concepts about oscillations and predict how variations in mass and spring characteristics can influence motion.

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Most popular questions from this chapter

On a smooth inclined plane, a body of mass \(M\) is attached between two springs. The other ends of the springs are fixed to firm support. If each spring has force constant \(k\), the period of oscillation of the body (assuming the springs as massless) is (a) \(2 \pi[M / 2 k]^{1 / 2}\) (b) \(2 \pi[2 M / k]^{1 / 2}\) (c) \(2 \pi[M g \sin \theta / 2 k]^{1 / 2}\) (d) \(2 \pi[2 M g / k]^{1 / 2}\)

A highly rigid cubical block \(A\) of small mass \(M\) and side \(L\) is fixed rigidly on the another cubical block of same dimensions and low modulus of rigidity \(\eta\) such that the lower face of \(A\) completely covers the upper face of \(B .\) The lower face of \(B\) is rigidly held on a horizontal surface. A small force \(F\) is applied perpendicular to one of the side faces of \(Z\). After the force is withdrawn, block \(A\) executes small oscillations, the time period of which is given by (a) \(2 \pi \sqrt{M L \eta}\) (b) \(2 \pi \sqrt{M \eta /} L\) (c) \(2 \pi \sqrt{M L / \eta}\) (d) \(2 \pi \sqrt{M / \eta} L\)

A simple pendulum of length \(l\) has been set up inside a railway wagon sliding down a frictionless inclined plane having an angle of inclination \(\theta=30^{\circ}\) with the horizontal. What will be its period of oscillation as recorded by an observer inside the wagon? (a) \(2 \pi \sqrt{\frac{2 l}{\sqrt{3} g}}\) (b) \(2 \pi \sqrt{2 l / g}\) (c) \(2 \pi \sqrt{l / g}\) (d) \(2 \pi \sqrt{\frac{\sqrt{3} l}{2 g}}\)

One end of a spring of force constant \(k\) is fixed to a vertical wall and the other to a block of mass \(m\) resting on a smooth horizontal surface. There is another wall at a distance \(x_{0}\) from the block. The spring is then compressed by \(2 x_{0}\) and then released. The time taken to strike the wall iv (a) \(\frac{1}{6} \pi \sqrt{\frac{k}{m}}\) (b) \(\sqrt{\frac{k}{m}}\) (c) \(\frac{2 \pi}{3} \sqrt{\frac{m}{k}}\) (d) \(\frac{\pi}{4} \sqrt{\frac{k}{m}}\)

One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. The suction pump is removed, the column of mercury in the U-tube will show (a) periodic motion (b) oscillation (c) simple harmonic motion (d) None of the above
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