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Chapter 16: Problem 56

56\. Vertical Spring A vertical spring stretches \(9.6 \mathrm{~cm}\) when a \(1.3 \mathrm{~kg}\) block is hung from its end. (a) Calculate the spring constant. This block is then displaced an additional \(5.0 \mathrm{~cm}\) downward and released from rest. Find (b) the period, (c) the frequency, (d) the amplitude, and (e) the maximum speed of the resulting SHM.

Short Answer

Expert verified
Spring constant: 132.3 N/m, Period: 0.63 s, Frequency: 1.59 Hz, Amplitude: 0.050 m, Maximum speed: 0.50 m/s.

Step by step solution

01

Calculate the Spring Constant

We need to use Hooke's Law, which is given by \[ F = kx \]. Here, the force (F) exerted by the block is the weight of the block, \[ F = mg \], where \( m = 1.3 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \).The displacement (x) is given as 9.6 cm, which converts to 0.096 m. Therefore, \[ mg = kx \ k = \frac{mg}{x} = \frac{1.3 \times 9.8}{0.096} = 132.3 \text{ N/m}. \]
02

Find the Period

The period (T) of simple harmonic motion is given by \[ T = 2\pi \sqrt{\frac{m}{k}}. \]Using the values \( m = 1.3 \text{ kg} \) and \( k = 132.3 \text{ N/m} \), we get \[ T = 2\pi \sqrt{\frac{1.3}{132.3}} \approx 0.63 \text{ s}. \]
03

Calculate the Frequency

The frequency (f) is the reciprocal of the period, \[ f = \frac{1}{T}. \]Using \( T \approx 0.63 \text{ s} \), we get \[ f \approx \frac{1}{0.63} \approx 1.59 \text{ Hz}. \]
04

Determine the Amplitude

The amplitude (A) is the maximum displacement from the equilibrium position. The block is displaced an additional \( 5.0 \text{ cm} \) downward, which is the amplitude. Therefore, \[ A = 5.0 \text{ cm} = 0.050 \text{ m}. \]
05

Find the Maximum Speed

The maximum speed (v_{max}) in simple harmonic motion is given by \[ v_{max} = A\omega, \]where \( \omega = 2\pi f \).Using \( A = 0.050 \text{ m} \) and\( f \approx 1.59 \text{ Hz} \), we first find \[ \omega = 2\pi \times 1.59 \approx 10 \text{ rad/s} \ v_{max} = 0.050 \times 10 \approx 0.50 \text{ m/s}. \]

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Key Concepts

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

Spring Constant Calculation
To find the spring constant, we use **Hooke's Law**, which states \( F = kx \). Here, **F** is the force exerted by the block due to its weight, calculated using **F = mg**. Given the block's mass (**m**) is 1.3 kg and gravity (**g**) is 9.8 m/s虏, the force is: \[ F = 1.3 \times 9.8 = 12.74 \text{ N} \] The spring stretches by 9.6 cm (0.096 m), so Hooke's Law gives: \[ k = \frac{F}{x} = \frac{12.74}{0.096} = 132.3 \text{ N/m} \] This is the spring constant.
Hooke's Law
Hooke's Law describes the relationship between the force exerted by a spring and its displacement. It鈥檚 written as: \[ F = kx \] where:
  • **F** is the force applied by the spring in Newtons (N)
  • **k** is the spring constant in N/m
  • **x** is the displacement from the equilibrium position in meters (m)
This linear relationship implies that the force needed to compress or extend a spring is directly proportional to the displacement.
Oscillation Period
The period (**T**) of simple harmonic motion is the time it takes for one complete oscillation. It鈥檚 determined by the formula: \[ T = 2\pi \sqrt{\frac{m}{k}} \] For our example, with mass **m** = 1.3 kg and spring constant **k** = 132.3 N/m, we find: \[ T = 2 \pi \sqrt{\frac{1.3}{132.3}} \approx 0.63 \text{ s} \] This value means the block completes one full oscillation every 0.63 seconds.
Frequency of Oscillation
The frequency (**f**) of oscillation tells us how many complete oscillations occur per second. It鈥檚 the reciprocal of the period and is calculated as: \[ f = \frac{1}{T} \] Using the period **T** 鈮� 0.63 s, we get: \[ f \approx \frac{1}{0.63} \approx 1.59 \text{ Hz} \] Thus, the block oscillates about 1.59 times per second.
Amplitude of Motion
The amplitude (**A**) is the maximum distance the block moves from its equilibrium position. In this example, the block is pulled an additional 5.0 cm downward before being released. Therefore, the amplitude is:
  • A = 5.0 cm = 0.050 m
This distance is the farthest point from the rest position the block reaches during its oscillation.
Maximum Speed
The maximum speed (**v_max**) of an oscillating object is reached when it passes through the equilibrium position. It's calculated using: \[ v_{max} = A\omega \] where **蠅** is the angular frequency, given by \[ \omega = 2\pi f \] With **f** 鈮� 1.59 Hz and **A** = 0.050 m, we find:
  • 蠅 鈮� 10 rad/s
Then, the maximum speed is: \[ v_{max} = 0.050 \times 10 = 0.50 \text{ m/s} \] This value shows the highest speed the block achieves during its movement.

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Most popular questions from this chapter

17\. Shake Table A block is on a horizontal surface (a shake table) that is moving back and forth horizontally with simple harmonic motion of frequency \(2.0 \mathrm{~Hz} .\) The coefficient of static friction between block and surface is \(0.50 .\) How great can the amplitude of the SHM be if the block is not to slip along the surface?

2\. Oscillating Block An oscillating block-spring system takes \(0.75 \mathrm{~s}\) to begin repeating its motion. Find its (a) period, (b) frequency in hertz, and (c) angular frequency in radians per second.

69\. To What Angle? A small metal ball of mass \(m\) hangs from a pivot by a rigid, light metal rod of length \(R .\) The ball is swinging back and forth with an amplitude that remains small throughout its motion, \(\theta^{\max } \leq 5^{\circ} .\) Ignore all damping. (a) The equation of motion of this ideal pendulum can be derived in a variety of ways and is $$ \frac{d^{2} \theta}{d t^{2}}=-\frac{g}{R} \sin \theta $$ For small angles, show how this can be replaced by an approximate equation of motion that can be solved more easily than the one given. (b) Write a general solution for the approximate equation of motion you obtained in (a) that works for any starting angle and angular velocity (as long as the angles stay in the range where the approximation is OK). Demonstrate that what you have written is a solution and show that at a time \(t=0\) your solution can have any given starting position and velocity. (c) If the length of the rod is \(0.3 \mathrm{~m}\), the mass of the ball is \(0.2 \mathrm{~kg}\), and the clock is started at a time when the ball is passing through the center \((\theta=0)\) and is moving with an angular speed of \(0.1 \mathrm{rad} / \mathrm{s}\), find the maximum angle your solution says the ball will reach. Can you use the approximate equation of motion for this motion? If the starting angle is not small, you cannot easily solve the equation of motion without a computer. But there are still things you can do. (d) Derive the energy conservation equation for the motion of the pendulum. (Do not use the small-amplitude approximation.) (e) If the pendulum is released from a starting angle of \(\theta_{1}\), what will be the maximum speed it travels at any point on its swing?

50\. Block-Spring An oscillating block-spring system has a mechanical energy of \(1.00 \mathrm{~J}\), an amplitude of \(10.0 \mathrm{~cm}\), and a maximum speed of \(1.20 \mathrm{~m} / \mathrm{s}\). Find (a) the spring constant, (b) the mass of the block, and (c) the frequency of oscillation.

58\. Lightly Damped The amplitude of a lightly damped oscillator decreases by \(3.0 \%\) during each cycle. What fraction of the mechanical energy of the oscillator is lost in each full oscillation?
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