Question

A body of mass \(1 \mathrm{~kg}\) suspended from the free end of a spring having force constant \(400 \mathrm{Nm}^{-1}\) is executing S.H.M. When the total energy of the system is 2 joule, the maximum acceleration is \(\ldots \ldots . \mathrm{ms}^{-2}\). (A) \(8 \mathrm{~ms}^{-2}\) (B) \(10 \mathrm{~ms}^{-2}\) (C) \(40 \mathrm{~ms}^{-2}\) (D) \(40 \mathrm{cms}^{-2}\)

Short answer

The maximum acceleration of the body is 40 m/s², which corresponds to option (C).

Step-by-step solution

Identify the given parameters

We are given the following parameters: Mass of the body (m) = 1 kg Force constant of the spring (k) = 400 Nm^{-1} Total mechanical energy of the system (E) = 2 J We are asked to find the maximum acceleration of the body.

Understanding energy conservation in SHM

In simple harmonic motion, the total mechanical energy of the system remains constant and is the sum of the kinetic energy (KE) and the potential energy (PE) stored in the spring. At the maximum displacement (amplitude, A) of the oscillations, the kinetic energy is zero, and all energy is in the form of potential energy stored in the spring. So, the total energy at maximum displacement can be expressed as: E = PE = \(\frac{1}{2}kA^2\)

Finding the amplitude

We have the total mechanical energy, and we can solve for the amplitude (A) using the conservation of mechanical energy: E = \(\frac{1}{2}kA^2\) 2 = \(\frac{1}{2}(400)A^2\) To find the amplitude, we can now solve for A: A^2 = \(\frac{2}{200}\) A = \(\sqrt{\frac{1}{100}}\) = 1/10 m

Finding the maximum acceleration

We can now find the maximum acceleration using the relationship between acceleration and displacement in simple harmonic motion: a = -k * x / m At the maximum displacement (x = A), the acceleration will be at its maximum value. Therefore: a_max = -k * A / m a_max = -400 * (1/10) / 1 a_max = -40 m/s^2 The maximum acceleration is negative because it acts in the opposite direction to the displacement. However, the magnitude of the maximum acceleration is: |a_max| = 40 m/s^2 So, the maximum acceleration of the body is 40 m/s², which corresponds to option (C).

Key concepts

Mechanical Energy Conservation

In simple harmonic motion (SHM), energy shifts between kinetic energy (KE) and potential energy (PE) while preserving the total mechanical energy. This conservation principle is what characterizes SHM as a continuous and smooth motion.
When a mass attached to a spring moves, it swings back and forth around an equilibrium point. At maximum displacement (either extreme), the energy is entirely potential, and kinetic energy is zero.
As the mass passes through the equilibrium position, the potential energy reaches a minimum while kinetic energy reaches a maximum. Regardless of these changes, the sum of KE and PE remains constant throughout the motion. This constancy is expressed mathematically as: \[ E = KE + PE = \text{constant} \]

Spring Constant

The spring constant, often denoted by \( k \), is a fundamental property of a spring and defines its stiffness. A stiffer spring requires more force to stretch or compress it by a given length compared to a less stiff spring. In SHM, the spring constant plays a crucial role in determining the motion's characteristics.
It is measured in newtons per meter (Nm⁻¹) and can be expressed through Hooke's Law, which states that the force \( F \) exerted by a spring is proportional to its displacement \( x \): \[ F = -kx \] This negative sign indicates that the force exerted by the spring is always in the opposite direction of displacement, showcasing the restoring nature of the force.
In problems involving SHM, such as the one a student faced in this case, knowing the spring constant allows for calculations of amplitude, frequency, and maximum acceleration.

Amplitude Determination

The amplitude \( A \) in simple harmonic motion is the maximum extent of displacement from the equilibrium position. It represents the peak distance that the mass on a spring or pendulum moves on either side of the equilibrium point.
When determining the amplitude through problem-solving, energy conservation principles are often employed. Here, the formula for total mechanical energy involves the potential energy at maximum displacement: \[ E = \frac{1}{2}kA^2 \]
By rearranging this equation, we can solve for \( A \): \[ A = \sqrt{\frac{2E}{k}} \] In this exercise, with a total energy of 2 joules and a spring constant of 400 Nm⁻¹, the amplitude was calculated by isolating \( A \) in the equation and finding its value as 0.1 meters.

Maximum Acceleration Calculation

The concept of maximum acceleration in simple harmonic motion is tied to the amplitude and the spring constant. It is the point where the restoring force is greatest, namely at the maximum displacement.
Using the basic relationship from SHM, acceleration \( a \) is derived from the spring constant \( k \), mass \( m \), and maximum displacement \( A \) by: \[ a = -\frac{k}{m} x \] For maximum acceleration, we substitute the maximum displacement \( x = A \): \[ a_{max} = -\frac{kA}{m} \]
Here, the acceleration is negative, indicating the direction of motion opposing the displacement. In magnitude, it becomes: \[ |a_{max}| = \frac{kA}{m} \] Substituting the provided values of \( k = 400 \,Nm^{-1} \), \( A = 0.1 \,m \), and \( m = 1 \,kg \), the calculation gives us the maximum acceleration of 40 m/s², showcasing the mechanical energy and forces at work in such harmonic systems.