Waves And Oscillations

and .
or or
It is an equation of a straight line
Therefore, the correct option is (C) : straight line

The correct option is (A) : T₁ < T₂

T =

case I : M₁ = 2M + M

case II : M₂ = 2M – M = = ⇒ T₂ = T₁

T is the oscillation period of a mass M hung from a light spring. The period of oscillation will now be if another mass M is suspended alongside it.

F ∝ X

Here, F = Force and X = Displacement

The period T of an oscillating mass M suspended from a light spring. Now, if another mass M is hanging beside it, the period of oscillation will be.

T = 2πM/K

The oscillating mass M's period T, which is hung from a light spring. Now, the period of oscillation will be if another mass M is hanging next to it.

T = 2πM/K

Now when two M masses are there then, M+M = 2M and now the time period is T₁

So, T₁ = 2π2M/K

T₁ = 2 x T = 2T

An oscillatory motion is simple harmonic motion (SHM). According to SHM, a particle's acceleration is shown to be inversely related to its displacement from the initial position at any given point. A specific type of oscillatory motion is known as SHM.

As a result, all simple harmonic movements are capable of having an oscillatory and periodic character. The opposite, however, is untrue. Not every oscillatory motion is an SHM.

Parameters in SHM that are affected by time include displacement, velocity, acceleration, and force. Sinusoids are sometimes referred to as sine, and the cosine functions serve to illustrate this. Simple Harmonic Motion explains the unique properties of alternating currents, sound waves, and light waves.

Solving the Equation for

We are given two equations involving and , and we need to solve for .

Step 1: Given Equations

The first equation is:

The second equation is:

Step 2: Taking the Ratio of the Equations

Now, we take the ratio of the two equations to eliminate . This gives:

Step 3: Squaring Both Sides

Next, we square both sides of the equation to remove the square root:

This simplifies to:

Step 4: Solving for

To solve for , we cross-multiply:

Now expand the equation:

Now, subtract from both sides:

Finally, solve for :

Conclusion:

The value of is .

To solve this problem, we need to determine the time period of oscillation for a mass suspended from a spring. The force required to stretch the spring is given, as well as the mass attached.

Start by understanding Hooke's Law, which states that the force needed to extend or compress a spring by distance is proportional to that distance. It can be described as:

Where:

• : Force applied to the spring.
• : Extension length of the spring, converted from centimeters to meters.
• is the spring constant we are looking to find.

Rearranging Hooke’s Law gives us:

Now, using the formula for the time period of a mass-spring system:

Where:

• : Mass attached to the spring.
• : Spring constant.

Substituting these values in gives:

Thus, the correct time period of oscillation is 0.628 seconds, which matches the given correct answer.

In simple harmonic motion (SHM), the potential energy (PE) of a body oscillating at a frequency depends on the displacement from its equilibrium position. The potential energy function is given by:

where is the spring constant and is the displacement.

The displacement varies as:

where:

• is the amplitude,
• is the angular frequency,
• is the phase constant.

The potential energy can be expressed as:

This can be rewritten in terms of trigonometric identities:

Thus, the expression becomes:

The term suggests that the potential energy oscillates with a frequency of .

Therefore, since , the frequency of the potential energy is .

Hence, the correct answer is 2n.

The correct answer is (a)

In a spring-mass system where a box filled with sand is attached to the spring, as sand leaks from the box, it results in a change in both the mass of the box and the spring dynamics. Let's analyze how this affects the average frequency (ω(t)) and amplitude (A(t)) of the oscillating system over time:

Frequency Analysis:

The natural frequency of a simple harmonic oscillator is given by:

ω = √(k/m)

where k is the spring constant and m is the mass of the system. As sand leaks out, the mass m decreases, leading to an increase in the frequency ω because the mass (m) is in the denominator. Thus, ω(t) increases over time.

Amplitude Analysis:

The amplitude of oscillation in a damped system can be influenced by external factors such as the loss of mass. When sand leaks out, the energy dissipation remains similar, but the effective damping due to air resistance relative to the mass increases, leading to a decrease in amplitude over time due to increased damping relative to reduced mass.

Therefore, A(t) decreases over time.

Conclusion:

The correct schematic for these changes would show an increase in frequency (ω) with time and a decrease in amplitude (A) with time. This is best represented by Figure 1, as it accurately depicts the described changes: increasing frequency and decreasing amplitude over time.