Question

A block with mass \(m =\) 0.300 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The other end of the spring is attached to a wall. When the block is at \(x = +\)0.240 m, its acceleration is \(a_x = -\)12.0 m/s\(^{2}\) and its velocity is \(v_x = +\)4.00 m/s. What are (a) the spring's force constant \(k\); (b) the amplitude of the motion; (c) the maximum speed of the block during its motion; and (d) the maximum magnitude of the block's acceleration during its motion?

Short answer

(a) 15.0 N/m; (b) 0.613 m; (c) 4.34 m/s; (d) 30.7 m/s².

Step-by-step solution

Calculating Spring Constant

To find the spring constant \(k\), we apply Hooke's law: \( F = -kx \). Here, the force \(F\) can be represented by the equation \(F = ma_x\), where \(m = 0.300\) kg and \(a_x = -12.0\) m/s\(^{2}\). Thus, \(F = 0.300 \times (-12.0) = -3.6\) N. Using Hooke's Law, we substitute into \(F = -kx\) to get \(-3.6 = -k(0.240)\). Solving for \(k\), we find \(k = \frac{3.6}{0.240} = 15.0\) N/m.

Finding the Amplitude of Motion

The amplitude \(A\) can be determined using the equation for the total energy in a spring system: \( E = \frac{1}{2} k A^2 = \frac{1}{2} k x^2 + \frac{1}{2} mv_x^2 \). Calculating the kinetic and potential energies: \( \frac{1}{2} k x^2 = \frac{1}{2} \times 15.0 \times (0.240)^2 = 0.432\) J and \( \frac{1}{2} mv_x^2 = \frac{1}{2} \times 0.300 \times (4.00)^2 = 2.4\) J. Total energy \( E = 0.432 + 2.4 = 2.832\) J. Solving \( \frac{1}{2} k A^2 = 2.832\), we find \( A = \sqrt{\frac{2 \times 2.832}{15.0}} = 0.613\) m.

Calculating the Maximum Speed

The maximum speed \(v_{max}\) occurs when all the energy is kinetic, \( E = \frac{1}{2} mv_{max}^2 \). Setting this equal to total energy, \( 2.832 = \frac{1}{2} \times 0.300 \times v_{max}^2 \). Solving for \(v_{max}\), \( v_{max} = \sqrt{\frac{2 \times 2.832}{0.300}} = 4.34\) m/s.

Finding the Maximum Acceleration

The maximum acceleration \( a_{max} \) occurs when the displacement is greatest (amplitude \( A \)), given by \( a_{max} = \frac{kA}{m} \). Substituting \( k = 15.0 \) N/m, \( A = 0.613 \) m, and \( m = 0.300 \) kg, we find \( a_{max} = \frac{15.0 \times 0.613}{0.300} = 30.7 \) m/s\(^{2}\).

Key concepts

Hooke's Law

In the world of physics, Hooke's Law describes how springs work. It states that the force needed to either compress or extend a spring is directly proportional to the distance the spring is stretched or compressed from its rest position. This can be summed up with the formula:

• Hooke's Law: \( F = -kx \)

Where:

• \( F \) is the force exerted by the spring,
• \( k \) is the spring constant,
• \( x \) is the displacement from the spring's equilibrium position.

The negative sign indicates that the force exerted by the spring is opposite to the direction of displacement, hence it is a restoring force that tries to bring the spring back to its original length.
When understanding simple harmonic motion, Hooke's Law is the foundational principle that describes how the mass-spring system oscillates.

Spring Constant

The spring constant, denoted as \( k \), measures a spring's stiffness. The higher the value, the stiffer the spring, meaning it requires more force to stretch or compress the spring. In the given exercise, we calculated the spring constant by applying Hooke’s Law:

• Force \( F \) is through the equation \( F = ma_x \),
• where \( m = 0.300 \) kg is the mass, and \( a_x = -12.0 \) m/s\(^2\) is the acceleration.

By substituting and solving, we found:

• \( k = \frac{3.6}{0.240} = 15.0 \) N/m

This tells us how strongly the spring resists deformation, a crucial factor in determining the characteristics of the motion.

Amplitude of Motion

In simple harmonic motion, the amplitude represents the maximum extent of the oscillation from the equilibrium position. Essentially, it’s the furthest point the mass will move from its starting point. To find the amplitude \( A \), we use the principle of energy conservation which states total mechanical energy is constant:

• Total energy \( E = \frac{1}{2} k A^2 = \frac{1}{2} k x^2 + \frac{1}{2} mv_x^2 \)

Here, the total energy is composed of both potential energy stored in the spring and the kinetic energy of the block. After calculations:

• Amplitude \( A = \sqrt{\frac{2 \times 2.832}{15.0}} = 0.613 \) m

This measure of amplitude indicates the greatest distance the mass moves from the equilibrium during its motion cycle.

Maximum Speed

The maximum speed is achieved when the oscillating mass passes through the equilibrium position, where all energy is kinetic. The kinetic energy is then at its peak since the potential energy in the spring is zero at this point. The formula to calculate this is derived from:

• \( E = \frac{1}{2} mv_{max}^2 \)

Substituting the total energy calculated earlier:

• \( v_{max} = \sqrt{\frac{2 \times 2.832}{0.300}} = 4.34 \) m/s

This formula directly ties the maximum speed to the conservation of energy, ensuring the energy is not lost but merely transferred between forms as the system oscillates.

Maximum Acceleration

The maximum acceleration occurs at the points of maximum displacement—the amplitude, where the restoring force exerted by the spring is greatest. To calculate maximum acceleration \( a_{max} \), we use:

• \( a_{max} = \frac{kA}{m} \)

Plugging in the values we calculated:

• \( a_{max} = \frac{15.0 \times 0.613}{0.300} = 30.7 \) m/s\(^2\)

This shows how the acceleration peaks at the endpoints of the spring's extension, emphasizing the relationship between spring constant, amplitude, and mass in determining dynamic properties of the motion.