Chapter 3: Problem 27
The number of oscillations completed in one second is called its frequency. The bob is oscillating once in every 6 seconds The number of oscillations in \(1 \mathrm{~s}=1 / 6\)
Chapter 3: Problem 27
The number of oscillations completed in one second is called its frequency. The bob is oscillating once in every 6 seconds The number of oscillations in \(1 \mathrm{~s}=1 / 6\)
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Get started for freeFill 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).
Speed \(=\frac{\text { Distance travelled }}{\text { Time taken }}\) \(\Rightarrow\) distance travelled by the body \(=\) speed of the body \(\times\) time taken.
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.
The rest and motion are relative. An object at rest with respect to one observer may not be at rest with respect to another observer. The same can be said about motion. For a person inside a bus, the fellow passengers are at rest but the same passengers are in motion with respect to a person standing on the ground.
From the given figure, by pythogorous theorem \(\mathrm{PR}^{2}=\mathrm{PQ}^{2}+\mathrm{R} \mathrm{Q}^{2}\) \(\mathrm{PR}^{2}=4^{2}+3^{2}\) \(\mathrm{PR}^{2}=16+9\) \(\mathrm{PR}=\sqrt{25}\) \(\mathrm{PR}=5 \mathrm{~m}\) Average speed \(=\frac{\text { total distance }}{\text { total time }}=\frac{4 \mathrm{~m}+3 \mathrm{~m}+5 \mathrm{~m}}{20 \mathrm{~s}+10 \mathrm{~s}+30 \mathrm{~s}}=\frac{12 \mathrm{~m}}{60 \mathrm{~s}}=0.2 \mathrm{~m} \mathrm{~s}^{-1}\)
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