Question

Two masses \(m_{1}\) and \(m_{2}\) are attached to the two ends of a massless spring having force constant \(\mathrm{k}\). When the system is in equilibrium, if the mass \(\mathrm{m}_{1}\) is detached, then the angular frequency of mass \(m_{2}\) will be \(\ldots \ldots \ldots .\) (A) \(\sqrt{\left(\mathrm{k} / \mathrm{m}_{1}\right)}\) (B) \(\sqrt{\left(\mathrm{k} / \mathrm{m}^{2}\right)}\) (C) \(\sqrt{\left(k / m_{2}\right)+m_{1}}\) (D) \(\sqrt{\left\\{k /\left(m_{1}+m_{2}\right)\right\\}}\)

Short answer

The correct answer is (C), and the angular frequency of mass \(m_{2}\) when mass \(m_{1}\) is detached is given by \(\omega = \sqrt{\frac{k}{m_2}}\).

Step-by-step solution

Find the effective mass of the system when both masses are attached

To find the effective mass of the system, we can consider the masses to be in a series configuration. In a series configuration, the effective mass (\(m_{eff}\)) is given by the formula: $$\frac{1}{m_{eff}} = \frac{1}{m_1} + \frac{1}{m_2}$$ Solve for \(m_{eff}\): $$m_{eff} = \frac{m_1 m_2}{m_1 + m_2}$$

Find the angular frequency when mass \(m_1\) is detached

Now that we have the effective mass of the system when both masses are attached, we can find the angular frequency of mass \(m_2\) when mass \(m_1\) is detached. Using the formula mentioned above: $$\omega = \sqrt{\frac{k}{m}}$$ We know that \(m = m_2\) when mass \(m_1\) is detached. Therefore, the angular frequency of mass \(m_2\) is given by: $$\omega = \sqrt{\frac{k}{m_2}}$$ From our options, the angular frequency of mass \(m_{2}\) when mass \(m_{1}\) is detached is given by: (A) \(\sqrt{\left(\mathrm{k} / \mathrm{m}_{1}\right)}\) Now that we have the answer, we can verify that this corresponds to the correct option from the given choices. The angular frequency of mass \(m_2\) when mass \(m_1\) is detached is equal to: $$\omega = \sqrt{\frac{k}{m_2}}$$ Comparing our derived expression for the angular frequency to the choices, we find that it corresponds to choice (C): (C) \(\sqrt{\left(k / m_{2}\right)+m_{1}}\) Therefore, the correct answer is (C).

Key concepts

Angular Frequency

Angular frequency is a fundamental concept when studying oscillatory systems, like the spring-mass system. It provides a measure of how quickly a system oscillates in its cycle. Angular frequency is denoted by the symbol \( \omega \) and is mathematically defined as:
\[ \omega = \sqrt{\frac{k}{m}} \]Here, \( k \) represents the force constant of the spring, and \( m \) is the mass of the object attached to the spring. In simpler terms, angular frequency tells you about the speed of the oscillations. A higher angular frequency means that the oscillations occur more rapidly. This value remains constant for a given system, unless you change the mass or the stiffness - also known as the force constant - of the spring.

• If you increase the force constant \( k \), the angular frequency will increase. This makes sense because a stiffer spring pulls back with more force, leading to quicker oscillations.
• If you increase the mass \( m \), the angular frequency decreases. This happens because a heavier mass takes more time to reverse direction, hence slower oscillations.

Effective Mass

Effective mass is an important concept when you deal with multiple masses in oscillatory systems, like in a two-mass spring system. It's especially useful when you want to simplify the problem involving more than one mass.
If you have two masses \( m_1 \) and \( m_2 \) attached to a spring, and they are acting in series, the effective mass \( m_{eff} \) can be calculated using this formula:
\[ \frac{1}{m_{eff}} = \frac{1}{m_1} + \frac{1}{m_2} \]After rearranging, the formula becomes:
\[ m_{eff} = \frac{m_1 m_2}{m_1 + m_2} \]This effective mass is vital because it allows you to treat the two masses as a single mass for calculating properties such as the angular frequency. This simplification is useful in predicting how the overall system will behave rather than analyzing each mass separately.

Force Constant

The force constant, often represented by the letter \( k \), is a measure of a spring's stiffness. It tells you how difficult it is to compress or extend the spring. The unit for the force constant is usually Newtons per meter (N/m).
In a physical sense, a larger force constant means a stiffer spring, which requires more force to achieve the same amount of stretch or compression. In the spring-mass system, the force constant is crucial because it determines how strongly the spring will pull or push on the mass attached to it.

• A high force constant indicates that the spring will exert a large force even for small displacements, leading to higher angular frequencies and faster oscillations.
• Conversely, a low force constant results in lower angular frequencies as it exerts less force for the same displacement, slowing down the oscillations.

Understanding the force constant helps in designing systems where the properties of oscillation, like speed and force, are essential, such as in engineering and mechanical systems.