Problem 20 - Master Resource Book in JEE Main Physics

Question

The frequency of vibration \(f\) of a mass \(m\) suspended from a spring of spring constant \(k\) is given by relation of the type \(f=c m^{x} k^{y}\), where \(c\) is a dimensionless constant. The values of \(x\) and \(y\) are (a) \(1 / 2,1 / 2\) (b) \(-1 / 2,-1 / 2\) (c) \(1 / 2,-1 / 2\) (d) \(-1 / 2,1 / 2\)

Step-by-Step Solution

Answer

(d) \(-1/2, 1/2\)

Step 1: Understand the Relationship

The given problem states that the frequency \( f \) is proportional to \( m^x \), the mass, and \( k^y \), the spring constant, with a constant \( c \). Mathematically, this is expressed as \( f = c \cdot m^x \cdot k^y \).

Step 2: Identify the Standard Formula for Frequency of a Spring System

From physics, the frequency \( f \) of a spring mass system is given by \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \). This means \( f = \frac{1}{2\pi} \cdot k^{1/2} \cdot m^{-1/2} \).

Step 3: Relate to the Proportionality Form

The equation from Step 2 simplifies to \( f = c \cdot k^{1/2} \cdot m^{-1/2} \) where \( c = \frac{1}{2\pi} \). This gives us the exponents for \( m \) and \( k \).

Step 4: Identify the Values of \( x \) and \( y \)

The exponents on \( m \) and \( k \) from Step 3 are \( x = -1/2 \) and \( y = 1/2 \).

Step 5: Match the Values with the Given Options

Compare the derived values \( x = -1/2 \) and \( y = 1/2 \) to the given options.

Key Concepts

Spring Constant

The spring constant, often denoted as \( k \), is a measure of a spring's stiffness. In other words, it tells us how much force is needed to stretch or compress the spring by a certain distance. The relationship between the force \( F \) applied to a spring and the displacement \( x \) experienced by the spring is given by Hooke's Law, expressed as \( F = kx \). The higher the spring constant, the stiffer the spring is, meaning more force is required to produce the same displacement.

• A stiff spring (large \( k \)) moves less when a force is applied.
• A loose spring (small \( k \)) moves significantly under the same force.

The spring constant is an essential factor in determining many properties of a mass-spring system, including the frequency at which it vibrates. Understanding this concept is crucial as it helps in analyzing how different springs will respond in varying conditions.

Mass-Spring System

In the world of physics, a mass-spring system is a classic model used to illustrate simple harmonic motion. This system consists of a mass attached to a spring, where the mass can move back and forth when displaced from its equilibrium position. The force that brings the mass back towards this position is provided by the spring itself. This force is directly proportional to the displacement of the mass and is explained through Hooke's Law.
The mass-spring system is pivotal in studying oscillatory motion. It can be used to:

• Model natural vibrations in mechanical systems.
• Analyze real-world scenarios like car suspensions and musical instruments.

By understanding how a mass-spring system works, you can predict how the system will behave under different forces, masses, or spring constants. The balance between the mass and the spring constant is what primarily determines how fast or slowly the system will vibrate.

Frequency of Vibration

The frequency of vibration in a mass-spring system is the number of complete oscillations or cycles the system makes in a unit of time. In our simple mass-spring system, the frequency \( f \) depends on both the mass \( m \) and the spring constant \( k \). Mathematically, it is given by the formula \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \). This shows that:

• Increasing the spring constant \( k \) will increase the frequency, resulting in faster oscillations.
• Increasing the mass \( m \) will decrease the frequency, causing slower oscillations.

Understanding the relationship between mass, spring constant, and frequency is crucial when designing systems that rely on specific vibrational properties, like clocks, seismographs, and certain electronic devices. By manipulating \( k \) and \( m \), you can fine-tune the system's natural frequency to suit various applications.