A point mass oscillates along the x-axis according to the law x = x_0 cos(omega t - pi/4). If the ac

Q: Why do we differentiate twice?

Because velocity is the rate of change of displacement ($dx/dt$), and acceleration is the rate of change of velocity ($dv/dt$). So, acceleration is the second derivative of displacement.

Q: Where does the $\omega^2$ come from?

Every time you differentiate the trigonometric function with respect to time, the chain rule requires you to multiply by the derivative of the inside function ($\omega t - \pi/4$), which is $\omega$. Doing this twice gives $\omega \times \omega = \omega^2$.

Q: Why can't the amplitude $A$ be negative?

By convention, the amplitude of a simple harmonic oscillator is defined as a positive quantity representing the maximum displacement from equilibrium. The negative sign indicates direction or phase, which is absorbed into the phase constant.

Q: How did $-\pi/4$ become $3\pi/4$?

We used the identity $-\cos(\theta) = \cos(\theta + \pi)$. Here, $\theta = \omega t - \pi/4$. Adding $\pi$ to $-\pi/4$ gives $-\pi/4 + 4\pi/4 = 3\pi/4$.

Q: What is the physical meaning of this phase difference?

A phase difference of $\pi$ (or $180^\circ$) between displacement and acceleration means they are always in opposite directions. When the particle is at maximum positive displacement, the acceleration is at maximum negative value, pulling it back to the center.

Question

A point mass oscillates along the x-axis according to the law $x = x_0 \cos(\omega t - \pi/4)$. If the acceleration of the particle is written as $a = A \cos(\omega t + \delta)$, then

Options:

(A) $A = x_0, \delta = -\frac{\pi}{4}$

(B) $A = x_0\omega^2, \delta = \frac{\pi}{4}$

(C) $A = x_0\omega^2, \delta = -\frac{\pi}{4}$

(D) $A = x_0\omega^2, \delta = \frac{3\pi}{4}$

Sigma Percentile · Accepted Answer

### The Starting Point: Displacement

Imagine a particle executing Simple Harmonic Motion (SHM) along the x-axis. Its position at any given time $t$ is perfectly described by a cosine wave. In our problem, the displacement equation is given as:

$$x = x_0 \cos\left(\omega t - \frac{\pi}{4}ight)$$

Here, $x_0$ represents the maximum displacement or the **amplitude** of the motion, $\omega$ is the angular frequency, and $-\frac{\pi}{4}$ is the initial phase constant. If we were to visualize this using a phasor diagram, at $t=0$, the displacement phasor would be pointing downwards at an angle of $45^\circ$ below the positive x-axis.

### The Calculus of Motion: Velocity and Acceleration

To find the acceleration of the particle, we must rely on the fundamental principles of kinematics. Acceleration is the rate of change of velocity, and velocity is the rate of change of displacement. Therefore, acceleration is the **second derivative** of displacement with respect to time.

Let's take the first derivative to find the velocity $v$:

$$v = \frac{dx}{dt} = \frac{d}{dt} \left[ x_0 \cos\left(\omega t - \frac{\pi}{4}ight) ight]$$

Using the chain rule, the derivative of $\cos(	heta)$ is $-\sin(	heta)$, and the derivative of the inner function $(\omega t - \pi/4)$ is $\omega$. This gives us:

$$v = -x_0 \omega \sin\left(\omega t - \frac{\pi}{4}ight)$$

Now, we differentiate the velocity to find the acceleration $a$:

$$a = \frac{dv}{dt} = \frac{d}{dt} \left[ -x_0 \omega \sin\left(\omega t - \frac{\pi}{4}ight) ight]$$

The derivative of $\sin(	heta)$ is $\cos(	heta)$, and once again, an $\omega$ pops out from the chain rule:

$$a = -x_0 \omega^2 \cos\left(\omega t - \frac{\pi}{4}ight)$$

### The Phase Shift: Aligning the Math

We have successfully derived the acceleration, but there is a catch. The problem asks us to compare our result with a standard form: $a = A \cos(\omega t + \delta)$. Notice that the amplitude $A$ in the standard form is a **positive** quantity, whereas our derived equation has a negative sign in front.

To fix this, we need to absorb the negative sign into the cosine function. We can use the trigonometric identity $-\cos(	heta) = \cos(	heta + \pi)$. By adding $\pi$ (or $180^\circ$) to the phase angle, we effectively flip the direction of the phasor, which is mathematically equivalent to multiplying by $-1$.

Let's apply this identity to our acceleration equation:

$$a = x_0 \omega^2 \cos\left(\omega t - \frac{\pi}{4} + \piight)$$

Simplifying the phase angle:

$$-\frac{\pi}{4} + \pi = -\frac{\pi}{4} + \frac{4\pi}{4} = \frac{3\pi}{4}$$

So, our final acceleration equation becomes:

$$a = x_0 \omega^2 \cos\left(\omega t + \frac{3\pi}{4}ight)$$

By directly comparing this with $a = A \cos(\omega t + \delta)$, we can clearly see that the amplitude $A = x_0 \omega^2$ and the phase constant $\delta = \frac{3\pi}{4}$. This perfectly matches option (d).

Animated Solution for Physics - Oscillations: A point mass oscillates along the x-axis according to the law $x = x_{0} cos (ω t - π /4)$ . If the acceleration of the particle is written as $a = A cos (ω t + δ)$ , then

Select Answer:

Visualized Solution

$x = x_{0} cos (ω t - π /4)$

$a = \frac{d ^{2} x}{d t ^{2}}$

$v = \frac{d x}{d t}$

$a = \frac{d v}{d t}$

$- cos (θ) = cos (θ + π)$

$A = x_{0} ω^{2}, δ = \frac{3 π}{4}$

$Phase of Velocity$

The Sigma Insight: Simple Harmonic Motion (SHM)

The Starting Point

The Calculus of Motion

The Phase Shift

Similar Questions

The $x - t$ graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at $t = 4/3 s$ is

A particle is making simple harmonic motion along the X-axis. If at a distances $x_{1}$ and $x_{2}$ from the mean position, the velocities of the particle are $v_{1}$ and $v_{2}$ respectively, then the time period of its oscillation is given as

The mass $M$ shown in the figure oscillates in simple harmonic motion with amplitude $A$ . The amplitude of the point $P$ is

A particle moves with simple harmonic motion in a straight line. In first $τ$ sec, after starting from rest, it travels a distance $a$ and in next $τ$ sec, it travels $2 a$ in same direction, then

A particle performs simple harmonic motion with amplitude $A$ . Its speed is trebled at the instant that it is at a distance $\frac{2}{3} A$ from equilibrium position. The new amplitude of the motion is

The displacement of a particle varies according to the relation $x = 4 (cos π t + sin π t)$ . The amplitude of the particle is

A particle of mass $m$ executes simple harmonic motion with amplitude $a$ and frequency $ν$ . The average kinetic energy during its motion from the position of equilibrium to the end is

A particle executes simple harmonic motion represented by displacement function as $x (t) = A sin (ω t + ϕ)$ If the position and velocity of the particle at $t = 0$ s are 2 cm and $2 ω$ cm s $^{- 1}$ respectively, then its amplitude is $x 2$ cm, where the value of $x$ is ............ .

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=x0​cos(ωt−π/4)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=dt2d2x​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=dtdx​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=dtdv​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } −cos(θ)=cos(θ+π)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A=x0​ω2,δ=43π​

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