/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 If \(x\) denotes displacement in... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If \(x\) denotes displacement in time \(t\) and \(x=a \cos t\), then the acceleration is a. \(a \cos t\) b. \(-a \cos t\) c. \(a \sin t\) d. \(-a \sin t\)

Short Answer

Expert verified
The acceleration is \(-a \cos t\), option (b).

Step by step solution

01

Understanding Displacement

The displacement of a body as a function of time is given as \( x = a \cos t \). This means the body's position changes in a cosine wave pattern with amplitude \( a \).
02

Finding Velocity

To find the velocity, which is the first derivative of displacement with respect to time, we differentiate \( x = a \cos t \) with respect to \( t \): \( v = \frac{d}{dt} a \cos t = -a \sin t \).
03

Finding Acceleration

Acceleration is the derivative of velocity with respect to time. We differentiate \( v = -a \sin t \) with respect to \( t \): \( a = \frac{d}{dt}(-a \sin t) = -a \cos t \).
04

Comparing with Options

Given the calculated acceleration \( -a \cos t \), we compare it to the provided options to find that the correct answer is option (b).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Displacement
In simple harmonic motion, displacement is a key concept that describes how far a body is from its equilibrium position as it oscillates. In the given function, \[ x = a \cos t \]"x" represents the displacement at any time \( t \). This equation tells us that the displacement varies in a cosine wave pattern. The letter "a" stands for the amplitude, which is the maximum distance the object moves from the center or equilibrium point.

Visualize it like a swinging pendulum, where the greatest height it reaches on either side is the amplitude. The cosine function indicates that the motion starts at the maximum displacement and passes through zero (equilibrium) to the maximum negative displacement, and this cycle repeats over time.
Finding Velocity
Velocity in physics is the rate of change of displacement with respect to time. It gives us an idea of how fast the object's position is changing, and in which direction. To find the velocity, we differentiate the displacement equation with respect to time. So, if we have:\[ x = a \cos t \]The velocity \( v \) is found by taking the derivative:\[ v = \frac{d}{dt} (a \cos t) = -a \sin t \]This means that the velocity follows a sine pattern, which is a cosine wave shifted by 90 degrees. The negative sign tells us that the direction of motion is opposite to that of the sine wave's increasing positive values.

Think about how, just after passing through the center position, the object moves fastest, both as it speeds up towards the extreme position and then slows down as it approaches it.
Determining Acceleration
Acceleration is about how the velocity of an object changes over time. In simple harmonic motion, acceleration is the second derivative of displacement, or the first derivative of velocity. Starting with our velocity function:\[ v = -a \sin t \]We find acceleration \( a \) by differentiating again:\[ a = \frac{d}{dt} (- a \sin t) = -a \cos t \]This result shows that the acceleration also varies in a pattern similar to displacement but often inversely. The acceleration is at its maximum when the object is at its extreme positions (where displacement is maximum) and zero at the equilibrium position.

Key points about Acceleration in Simple Harmonic Motion:
  • It always acts towards the center, trying to restore the object to equilibrium.
  • The magnitude of acceleration is proportional to the displacement.
  • It contributes to the oscillating nature of the motion.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A stone is dropped from the top of a tower of height \(h\). After \(1 \mathrm{~s}\) another stone is dropped from the balcony \(20 \mathrm{~m}\) below the top. Both reach the bottom simultaneously. What is the value of \(h\) ? Take \(g=10 \mathrm{~ms}^{-2}\). a. \(3125 \mathrm{~m}\) b. \(312.5 \mathrm{~m}\) c. \(31.25 \mathrm{~m}\) d. \(25.31 \mathrm{~m}\)

When the speed of a car is \(u\), the minimum distance over which it can be stopped is \(s .\) If the speed becomes \(n u\), what will be the minimum distance over which it can be stopped during the same time? a. \(s / n\) b. \(n s\) c. \(s / n^{2}\) d. \(n^{2} s\)

The position \(x\) of a particle varies with time \((t)\) as \(x=a t^{2}-b t^{3} .\) The acceleration at time \(t\) of the particle will be equal to zero, where \(t\) is equal to a. \(\frac{2 a}{3 b}\) b. \(\frac{a}{b}\) c. \(\frac{a}{3 b}\) d. zero

A body is dropped from a height of \(39.2 \mathrm{~m}\). After it crosses half distance, the acceleration due to gravity ceases to act. The body will hit the ground with velocity (Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) ) a. \(19.6 \mathrm{~m} / \mathrm{s}\) b. \(20 \mathrm{~m} / \mathrm{s}\) c. \(1.96 \mathrm{~m} / \mathrm{s}\) d. \(196 \mathrm{~m} / \mathrm{s}\)

A ball is released from the top of a tower of height \(h\). It takes time \(T\) to reach the ground. What is the position of the ball (from ground) after time \(T / 3 ?\) a. \(h / 9 \mathrm{~m}\) b. \(7 h / 9 \mathrm{~m}\) c. \(8 h / 9 \mathrm{~m}\) d. \(17 h / 18 \mathrm{~m}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.