The displacement of an object attached to a spring and executing simple harmonic motion is given by

Q: Why does the object start at the extreme position?

Because the displacement is given by a cosine function. At $t = 0$, $\cos(0) = 1$, which means the displacement is at its maximum positive value, $x = A$.

Q: Why is the speed maximum at the mean position?

At the extreme position, the spring is fully stretched, and all energy is potential. As it moves towards the center, this potential energy converts into kinetic energy. At the mean position, the spring is relaxed, potential energy is zero, and kinetic energy (and thus speed) is at its absolute maximum.

Q: What if the equation was a sine function instead?

If it were $x = A \sin(\omega t)$, the object would start at the mean position at $t = 0$. In that case, the maximum speed would occur right at the start ($t = 0$), and then again after every half time period ($T/2$).

Q: How do we know it's the *first* time the maximum speed occurs?

The object starts at the extreme position and moves towards the mean position. The very first time it reaches the mean position is after completing one-fourth of its full oscillation cycle, which takes a time of $T/4$.

Q: Does the amplitude affect the time taken to reach the mean position?

No, it doesn't! One of the beautiful properties of Simple Harmonic Motion is that the time period is completely independent of the amplitude. Whether you pull the spring $2 \text{ cm}$ or $10 \text{ cm}$, it will take the exact same time to reach the center.

Question

The displacement of an object attached to a spring and executing simple harmonic motion is given by $x = 2 	imes 10^{-2} \cos \pi t$ metre. The time at which the maximum speed first occurs is

Options:

(A) $0.5 	ext{ s}$

(B) $0.75 	ext{ s}$

(C) $0.125 	ext{ s}$

(D) $0.25 	ext{ s}$

Sigma Percentile · Accepted Answer

### Decoding the Motion Equation

Imagine a block attached to a spring, resting on a frictionless table. When you pull the block and let it go, it begins to oscillate back and forth. This is the essence of Simple Harmonic Motion (SHM). In our problem, the displacement of the object is governed by the equation:

$$x = 2 	imes 10^{-2} \cos(\pi t)$$

The first thing we need to do is understand where the object is at the very beginning. If we plug in $t = 0$ into our equation, the cosine term becomes $\cos(0)$, which is exactly $1$. This tells us that at the start of our stopwatch, the object is at $x = 2 	imes 10^{-2} 	ext{ m}$. This is its maximum positive displacement, also known as the **positive extreme position**.

### The Physics of Maximum Speed

Now, the question asks us to find the time when the object first reaches its **maximum speed**. Let's visualize the physics here. When the block is at the extreme position, the spring is fully stretched. For a brief moment, the block stops before reversing direction. At this point, its speed is zero, and all its energy is stored as potential energy in the spring.

As the spring pulls the block back towards the center, it accelerates. The potential energy is rapidly converting into kinetic energy. The block will move the fastest exactly at the **mean position** ($x = 0$), where the spring is completely relaxed and neither pushing nor pulling.

Since the block starts at the extreme position and needs to reach the mean position, it has to complete exactly **one-fourth** of its full oscillation cycle. Therefore, the time required is simply:

$$t = \frac{T}{4}$$

### Calculating the Time

To find the time period $T$, we compare our given equation with the standard general equation for SHM:

$$x = A \cos(\omega t)$$

By comparing the terms, we can easily spot that the angular frequency $\omega$ is the coefficient of $t$. So, $\omega = \pi 	ext{ rad/s}$. The relationship between the time period and angular frequency is given by the formula:

$$T = \frac{2\pi}{\omega}$$

Substituting our value of $\omega$, we get:

$$T = \frac{2\pi}{\pi} = 2 	ext{ s}$$

This means the block takes exactly $2 	ext{ seconds}$ to complete one full back-and-forth cycle. Finally, to find the time it takes to reach the mean position for the first time, we divide this time period by four:

$$t = \frac{2}{4} = 0.5 	ext{ s}$$

And there we have it! Exactly half a second after release, the object will zip through the center point at its absolute maximum speed.

Animated Solution for Physics - Oscillations: The displacement of an object attached to a spring and executing simple harmonic motion is given by $x = 2 \times 1 0^{- 2} cos π t$ metre. The time at which the maximum speed first occurs is

Select Answer:

Visualized Solution

Equation of SHM $x = A cos (ω t)$

Maximum Speed at $x = 0$

Time to Mean Position $t = \frac{T}{4}$

Finding Time Period $T$

Final Calculation for $t$

Subsequent $v_{m a x}$

The Sigma Insight: Simple Harmonic Motion (SHM)

Decoding the Motion Equation

The Physics of Maximum Speed

Calculating the Time

Similar Questions

The maximum velocity of a particle, executing simple harmonic motion with an amplitude $7 mm$ , is $4.4 ms^{- 1}$ . The period of oscillation 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 ............ .

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

If a simple harmonic motion is represented by $\frac{d ^{2} x}{d t ^{2}} + α x = 0$ , its time period is

A simple harmonic motion is represented by $y = 5 (sin 3 π t + 3 cos 3 π t)$ cm. The amplitude and time period of the motion are

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 executes simple harmonic motion between $x = - A$ and $x = + A$ . The time taken for it to go from $O$ to $A /2$ is $T_{1}$ and to go from $A /2$ to $A$ is $T_{2}$ , then

Select Answer:

Visualized Solution

Equation of SHM .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=Acos(ωt)

Maximum Speed at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Time to Mean Position .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=4T​

Finding Time Period .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

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

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