The motion of a mass on a spring, with spring constant k is as shown in figure. The equation of moti

Q: Why do we multiply and divide by $\sqrt{A^2 + B^2}$?

This is a standard mathematical technique to combine a linear combination of sine and cosine into a single trigonometric function. It allows us to define a right-angled triangle where the coefficients act as the base and perpendicular, making it easy to use angle addition formulas.

Q: How did we get the velocity equation?

Velocity is the rate of change of displacement. By differentiating the position equation $x(t) = A \sin \omega t + B \cos \omega t$ with respect to time $t$, we get $v(t) = A\omega \cos \omega t - B\omega \sin \omega t$.

Q: Why did we choose $\cos(\omega t - \phi)$ instead of $\sin(\omega t + \phi)$?

The problem specifically asks to represent the displacement in the form $C \cos(\omega t - \phi)$. We tailored our trigonometric substitutions ($\sin \phi$ and $\cos \phi$) to match the expansion formula for $\cos(X - Y)$.

Q: What is the physical significance of $C$ and $\phi$?

$C$ represents the resultant amplitude of the simple harmonic motion, which is the maximum displacement from the mean position. $\phi$ is the initial phase angle, which determines the starting position of the mass at $t=0$.

Q: Can we solve this using phasors?

Absolutely! The terms $A \sin \omega t$ and $B \cos \omega t$ can be represented as two perpendicular phasors of lengths $A$ and $B$. Their vector sum gives the resultant phasor of length $C = \sqrt{A^2 + B^2}$ at an angle $\phi$.

Question

The motion of a mass on a spring, with spring constant $k$ is as shown in figure.

The equation of motion is given by $x(t) = A \sin \omega t + B \cos \omega t$ with $\omega = \sqrt{\frac{k}{m}}$.

Suppose that at time $t = 0$, the position of mass is $x(0)$ and velocity $v(0)$, then its displacement can also be represented as $x(t) = C \cos(\omega t - \phi)$, where $C$ and $\phi$ are

Options:

(A) $C = \sqrt{\frac{2v(0)^2}{\omega^2} + x(0)^2}, \phi = 	an^{-1}\left(\frac{v(0)}{x(0)\omega}ight)$

(B) $C = \sqrt{\frac{2v(0)^2}{\omega^2} + x(0)^2}, \phi = 	an^{-1}\left(\frac{x(0)\omega}{2v(0)}ight)$

(C) $C = \sqrt{\frac{v(0)^2}{\omega^2} + x(0)^2}, \phi = 	an^{-1}\left(\frac{x(0)\omega}{v(0)}ight)$

(D) $C = \sqrt{\frac{v(0)^2}{\omega^2} + x(0)^2}, \phi = 	an^{-1}\left(\frac{v(0)}{x(0)\omega}ight)$

Sigma Percentile · Accepted Answer

Combining trigonometric functions is a powerful technique in physics, especially when dealing with Simple Harmonic Motion (SHM). In this problem, we are given the displacement of a mass-spring system as a sum of sine and cosine functions, and we need to express it as a single cosine function. Let's dive into the elegant math behind this transformation!

### The Mathematical Trick

We start with the given equation of motion:
$$x(t) = A \sin \omega t + B \cos \omega t$$

Our goal is to mold this into the form $x(t) = C \cos(\omega t - \phi)$. To achieve this, we use a classic mathematical trick. We multiply and divide the entire expression by the square root of the sum of the squares of the coefficients, which is $\sqrt{A^2 + B^2}$:
$$x(t) = \sqrt{A^2 + B^2} \left[ \frac{A}{\sqrt{A^2 + B^2}} \sin \omega t + \frac{B}{\sqrt{A^2 + B^2}} \cos \omega t ight]$$

### The Phasor Triangle

Now, imagine a right-angled triangle where the base is $B$ and the perpendicular is $A$. By the Pythagorean theorem, the hypotenuse is $\sqrt{A^2 + B^2}$. We can define an angle $\phi$ such that:
$$\sin \phi = \frac{A}{\sqrt{A^2 + B^2}}$$
$$\cos \phi = \frac{B}{\sqrt{A^2 + B^2}}$$

From this, it naturally follows that:
$$	an \phi = \frac{A}{B}$$

Substituting these trigonometric ratios back into our equation, we get:
$$x(t) = \sqrt{A^2 + B^2} (\sin \phi \sin \omega t + \cos \phi \cos \omega t)$$

Using the trigonometric identity $\cos(X - Y) = \cos X \cos Y + \sin X \sin Y$, this perfectly simplifies to:
$$x(t) = \sqrt{A^2 + B^2} \cos(\omega t - \phi)$$

Comparing this with our target equation $x(t) = C \cos(\omega t - \phi)$, we immediately see that the new amplitude is:
$$C = \sqrt{A^2 + B^2}$$

### Applying Initial Conditions

Now we need to find the values of $A$ and $B$ in terms of the initial conditions provided in the problem.

**For Position:**
At $t = 0$, the position is $x(0)$. Plugging $t = 0$ into our original equation:
$$x(0) = A \sin(0) + B \cos(0)$$
Since $\sin(0) = 0$ and $\cos(0) = 1$, we get:
$$B = x(0)$$

**For Velocity:**
Velocity is the derivative of position with respect to time. Differentiating $x(t)$ gives:
$$v(t) = \frac{dx}{dt} = A\omega \cos \omega t - B\omega \sin \omega t$$

At $t = 0$, the velocity is $v(0)$. Plugging $t = 0$ into the velocity equation:
$$v(0) = A\omega \cos(0) - B\omega \sin(0)$$
$$v(0) = A\omega$$
$$A = \frac{v(0)}{\omega}$$

### The Grand Finale

Finally, let's substitute the values of $A$ and $B$ back into our formulas for $C$ and $	an \phi$.

For the amplitude $C$:
$$C = \sqrt{A^2 + B^2} = \sqrt{\left(\frac{v(0)}{\omega}ight)^2 + x(0)^2} = \sqrt{\frac{v(0)^2}{\omega^2} + x(0)^2}$$

For the phase constant $\phi$:
$$	an \phi = \frac{A}{B} = \frac{v(0)/\omega}{x(0)} = \frac{v(0)}{x(0)\omega}$$
$$\phi = 	an^{-1}\left(\frac{v(0)}{x(0)\omega}ight)$$

And there we have it! The displacement is beautifully represented as a single cosine function with the calculated amplitude and phase.

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Visualized Solution

$x (t) = A sin ω t + B cos ω t$

$Mathematical Transformation$

$Defining the Phase Angle$

$The Single Cosine Form$

$Applying Initial Conditions for Position$

$Applying Initial Conditions for Velocity$

$Final Substitution$

The Sigma Insight: Simple Harmonic Motion (SHM)

The Mathematical Trick

The Phasor Triangle

Applying Initial Conditions

The Grand Finale

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x(t)=Asinωt+Bcosωt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mathematical Transformation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Defining the Phase Angle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Single Cosine Form

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying Initial Conditions for Position

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying Initial Conditions for Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Substitution