/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 \(\mathrm{A}\) block with mass \... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\mathrm{A}\) block with mass \(m\) is undergoing SHM on a horizontal, frictionless surface while attached to a light, horizontal spring that has force constant \(k\). You use motion sensor equipment to measure the maximum speed of the block during its oscillations. You repeat the measurement for the same spring and blocks of different masses while keeping the amplitude \(A\) at a constant value of \(12.0 \mathrm{~cm}\). You plot your data as \(v_{\max }^{2}\) versus \(1 / m\) and find that the data lie close to a straight line that has slope \(8.62 \mathrm{~N} \cdot \mathrm{m} .\) What is the force constant \(k\) of the spring?

Short Answer

Expert verified
The force constant \(k\) of the spring is approximately \(600 \mathrm{N/m}\)

Step by step solution

01

Write down known values

The amplitude \(A\) of the oscillation is given as \(12.0 \mathrm{cm}\) or \(0.12 \mathrm{m}\), and the slope of the graph \(m\) is given as \(8.62 \mathrm{N} \cdot \mathrm{m}\). The slope of the graph corresponds to the value of \(kA^{2}\).
02

Calculate the spring constant

Knowing that \(kA^{2}\) is equal to the slope \(m = 8.62 \mathrm{N} \cdot \mathrm{m}\), we solve for \(k\), the spring constant. The formula for the spring constant \(k\) is obtained by rearranging \(kA^{2} = m\) to get \(k = \frac{m}{A^{2}}\). Substituting the known values gives, \(k = \frac{8.62 \mathrm{N} \cdot \mathrm{m}}{(0.12 \mathrm{~m})^2}\)
03

Simplify and solve

Doing the calculation we get \(k = 599.3 \mathrm{N/m}\). So, the spring constant \(k\) is approximately \(600 \mathrm{N/m}\).

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Key Concepts

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

Spring Constant
In simple harmonic motion (SHM), the spring constant, often denoted as \( k \), is a crucial parameter. It defines how stiff a spring is and determines the force required to stretch or compress it by a given distance. For springs, Hooke's Law states that this force \( F \) is proportional to the displacement \( x \) from its equilibrium position, mathematically given by \( F = -kx \). The negative sign indicates that the force is opposite to the displacement direction, acting to restore equilibrium.
In our exercise, we use the relationship between the maximum speed of an oscillating block and the mass to identify the spring constant. This is achieved by plotting the square of the maximum speed against the reciprocal of mass, where the slope of this line directly allows us to calculate \( k \) using the formula \( k = \frac{m}{A^2} \). This setup highlights how small deviations and relationships in SHM can be used to extract meaningful physical properties of the system.
Understanding the spring constant is fundamental, especially in physics and engineering, as it affects dynamics and energy storage in spring-mass systems.
Maximum Speed of Oscillation
The maximum speed of oscillation in simple harmonic motion is the peak speed an object attains during its motion. In our given setup, the object is a block oscillating on a frictionless surface attached to a spring. The maximum speed \( v_{max} \) of the block can be derived using the relationship \( v_{max} = A\omega \), where \( A \) is the amplitude, and \( \omega \) is the angular frequency. Angular frequency itself is calculated using the formula \( \omega = \sqrt{\frac{k}{m}} \).
The exercise effectively exemplifies how changes in mass, while keeping the spring constant and amplitude steady, influence \( v_{max} \). Observing \( v_{max}^2 \) versus \( 1/m \) draws a linear relationship, offering insights into the spring's force constant directly. Such exercises help in deepening the understanding of energy transfer, where the kinetic energy is maximum when all potential energy in the spring converts during equilibrium.
Motion Sensor Equipment
Motion sensor equipment plays an integral role in measuring various aspects of motion, including position, velocity, and acceleration. In our scenario, a motion sensor tracks the block's oscillation to measure its maximum speed accurately. These sensors typically employ technologies like infrared, ultrasound, or laser to determine object movement in real-time.
By capturing detailed data, the sensors help plot graphs like \( v_{max}^2 \) against \( 1/m \), which are instrumental in understanding the dynamics of the oscillations. This application underlines the importance of accurate data collection in physics experiments, enabling precise quantification of parameters like the spring constant, and providing an empirical foundation to theoretical models.
  • Enhances data collection accuracy
  • Facilitates visualization of motion dynamics
  • Supports empirical validation of theoretical concepts
Motion sensors bridge practical experimentation with theoretical insights, offering powerful insights into both the analyzed physical systems and the nature of simple harmonic motion itself.
Mass and Amplitude Relationship
The interaction between mass and amplitude in simple harmonic motion is an intriguing aspect of SHM systems. While amplitude \( A \) is typically a measure of the maximum extent of displacement from equilibrium, it remains constant in our experiment. Hence, the focus shifts to how variations in mass \( m \) affect the system.
In SHM, although the amplitude depends on initial conditions, the mass primarily influences the system's frequency and period. The period \( T \) of oscillation, \( T = 2\pi \sqrt{\frac{m}{k}} \), indicates that more massive objects will generally oscillate slower than lighter ones when attached to the same spring.
By keeping \( A \) constant and varying \( m \), the exercise accentuates how these two parameters independently exert influence. However, the maximum speed is a function of both \( m \) and \( k \), as shown in the calculated slopes, reflecting SHM's intricate balance of energy distribution between kinetic and potential forms. Such relationships are key in understanding harmonic oscillators in more complex systems, ranging from mechanical structures to atomic particles.

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Most popular questions from this chapter

A Spring with Mass. The preceding problems in this chapter have assumed that the springs had negligible mass. But of course no spring is completely massless. To find the effect of the spring's mass, consider a spring with mass \(M,\) equilibrium length \(L_{0},\) and spring constant \(k\). When stretched or compressed to a length \(L,\) the potential energy is \(\frac{1}{2} k x^{2},\) where \(x=L-L_{0}\). (a) Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v\). Assume that the speed of points along the length of the spring varies linearly with distance \(l\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. Calculate the kinetic energy of the spring in terms of \(M\) and \(v .\) (Hint: Divide the spring into pieces of length \(d l ;\) find the speed of each piece in terms of \(l, v,\) and \(L ;\) find the mass of each piece in terms of \(d l, M,\) and \(L ;\) and integrate from 0 to \(L .\) The result is \(n o t \frac{1}{2} M v^{2},\) since not all of the spring moves with the same speed.) (b) Take the time derivative of the conservation of energy equation, Eq. (14.21), for a mass \(m\) moving on the end of a massless spring. By comparing your results to Eq. (14.8), which defines \(\omega\), show that the angular frequency of oscillation is \(\omega=\sqrt{k / m}\). (c) Apply the procedure of part (b) to obtain the angular frequency of oscillation \(\omega\) of the spring considered in part (a). If the effective mass \(M^{\prime}\) of the spring is defined by \(\omega=\sqrt{k / M^{\prime}},\) what is \(M^{\prime}\) in terms of \(M ?\)

A \(40.0 \mathrm{~N}\) force stretches a vertical spring \(0.250 \mathrm{~m}\). (a) What mass must be suspended from the spring so that the system will oscillate with a period of \(1.00 \mathrm{~s} ?\) (b) If the amplitude of the motion is \(0.050 \mathrm{~m}\) and the period is that specified in part (a), where is the object and in what direction is it moving \(0.35 \mathrm{~s}\) after it has passed the equilibrium position, moving downward? (c) What force (magnitude and direction) does the spring exert on the object when it is \(0.030 \mathrm{~m}\) below the equilibrium position, moving upward?

A block of mass \(m\) is undergoing SHM on a horizontal, frictionless surface while attached to a light, horizontal spring. The spring has force constant \(k\), and the amplitude of the \(\mathrm{SHM}\) is \(A\). The block has \(v=0,\) and \(x=+A\) at \(t=0 .\) It first reaches \(x=0\) when \(t=T / 4\) where \(T\) is the period of the motion. (a) In terms of \(T,\) what is the time \(t\) when the block first reaches \(x=A / 2 ?\) (b) The block has its maximum speed when \(t=T / 4\). What is the value of \(t\) when the speed of the block first reaches the value \(v_{\max } / 2 ?\) (c) Does \(v=v_{\max } / 2\) when \(x=A / 2 ?\)

\(\mathrm{A}\) mass is oscillating with amplitude \(A\) at the end of a spring. How far (in terms of \(A\) ) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy?

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length \(50.0 \mathrm{~cm} .\) She finds that the pendulum makes 100 complete swings in 136 s. What is the value of \(g\) on this planet?

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