Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 8

Fill in the Blanks. \(\frac{\mathrm{T}}{2}\) The time taken to perform one to-and-fro motion (or) from one extreme position to other extreme position and back is called time period (T).

Short Answer

Expert verified
Question: In a to-and-fro motion, the time taken to perform half a cycle, such as moving from one extreme position to the other extreme position, is represented by the fraction ______. Answer: \(\frac{\mathrm{T}}{2}\)

Step by step solution

01

Understand Time Period

The time period (T) for a to-and-fro motion represents the amount of time it takes for an object (such as a pendulum) to make one complete cycle, starting and ending at the same point in its journey. This includes moving away from the starting point to its extreme position, then back towards and passing the starting point, until it reaches the other extreme position, and then returning to the starting point again.
02

Relate to the given fraction

Now that the concept of the time period is understood, let's focus on the given fraction \(\frac{\mathrm{T}}{2}\). This represents half of the time period (T). It means that this amount of time corresponds to half a cycle of the motion.
03

Fill in the Blank

As the exercise asks to fill in the blank to understand the given fraction, we can mention that \(\frac{\mathrm{T}}{2}\) corresponds to the time taken to perform half of a to-and-fro motion, i.e., moving from one extreme position to the other extreme position (or) from the starting point to the other extreme position without returning.
04

Summary

In summary, the given fraction \(\frac{\mathrm{T}}{2}\) denotes the time taken to perform half a to-and-fro motion, which can be understood as either moving from one extreme position to the other extreme position or moving from the starting point to the other extreme position without returning. This understanding helps to fill in the blank and ensures a clear idea of how the fraction relates to the time period (T) of a to-and-fro motion.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Odometer is used to find the distance travelled by the vehicle and speedometer is used to find the speed of the vehicle.

To mark a point for set of values \((1,5)\), look for 1 s on X-axis. Draw a line parallel to Y-axis and passing through this point. Look for \(5 \mathrm{~m}\) on the Y-axis and draw a line parallel to the X-axis passing through this point. The point of intersection gives the point that represents \((1,5) .\) In the same way, the points for other set of values can be plotted, as shown in the figure. (i) Join all the points. It is a straight line. The straight line is the distance-time graph for the motion of the motorbike. (ii) The distance-time graph of a body moving with a constant speed is a straight line. However, if the body does not move with constant speed, then its distance-time graph cannot be a straight line.

$$ \begin{array}{l} \mathrm{V}_{\mathrm{A}}=\frac{\mathrm{d}_{\mathrm{A}}}{\mathrm{t}} ; \mathrm{V}_{\mathrm{B}}=\frac{\mathrm{d}_{\mathrm{B}}}{\mathrm{t}} \\ \mathrm{d}_{\mathrm{A}}=80 \mathrm{~m} \text { and } \mathrm{d}_{\mathrm{B}}=100 \mathrm{~m} \\ \mathrm{~V}_{\mathrm{A}}=\frac{80}{\mathrm{t}} ; \mathrm{V}_{\mathrm{B}}=\frac{100}{\mathrm{t}} \Rightarrow \mathrm{V}_{\mathrm{A}}<\mathrm{V}_{\mathrm{B}} \cdot \mathrm{B}^{\prime} \text { is faster than }^{4} \mathrm{~A}^{\prime} . \end{array} $$

Fill in the Blanks. vibratory The molecules in solid undergo vibratory motion.

(i) The unit of speed and velocity is \(\mathrm{m} \mathrm{s}^{-1}\). (ii) If a body is moving in a straight line path then its speed is equal to velocity \(50 \mathrm{~m} \mathrm{~s}^{-1}=50 \times \frac{18}{5} \mathrm{~km} \mathrm{~h}^{-1}=180 \mathrm{~km} \mathrm{~h}^{-1}\)
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Wave Optics

Read Explanation

Oscillations

Read Explanation
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.

Sign-up for free