1. The term oscillation most nearly refers to a motion in which a body
ⓐ. moves once from one place to another and then stops
ⓑ. moves only in a circular path with constant speed
ⓒ. keeps increasing its distance from a fixed point
ⓓ. repeats about an equilibrium or mean position
Correct Answer: repeats about an equilibrium or mean position
Explanation: Oscillation means repeated motion about an equilibrium or mean position. A swing, a simple pendulum, and a mass attached to a spring can all move to and fro about a central position. The body need not move along a straight line in every example, but there must be repeated departure from and return toward the mean position. Motion that simply continues in one direction is not oscillation, even if it is smooth. The central idea is repeated motion with a tendency to come back toward a reference position.
2. A motion repeats its state after equal intervals of time, but it does not move to and fro about a mean position. This motion is best described as
ⓐ. oscillatory but not periodic
ⓑ. simple harmonic by definition
ⓒ. neither periodic nor repeated
ⓓ. periodic but not oscillatory
Correct Answer: periodic but not oscillatory
Explanation: A periodic motion repeats itself after equal intervals of time. Oscillatory motion is more specific because it must be to-and-fro motion about an equilibrium or mean position. Uniform circular motion is a common example of periodic motion that is not normally called oscillatory motion of the body itself. Every oscillatory motion is periodic only when it repeats regularly, but every periodic motion is not necessarily oscillatory. The phrase “to and fro about a mean position” is the extra condition needed for oscillation.
3. In a playground swing, the lowest position of the seat is usually treated as the mean position. When the swing is pulled aside and released, why does it come back toward this position?
ⓐ. The amplitude becomes zero at the extreme position
ⓑ. A restoring tendency acts toward the mean position
ⓒ. The time period changes direction at the mean position
ⓓ. The frequency becomes larger near the extreme position
Correct Answer: A restoring tendency acts toward the mean position
Explanation: A swing displaced from its lowest position tends to return toward that position. This return is due to a restoring tendency that acts toward the mean or equilibrium position. The swing does not stop permanently at the mean position because it has inertia and can overshoot to the other side. Amplitude tells how far the swing goes from the mean position, not why it returns. The restoring tendency and inertia together produce the repeated motion.
4. A vibrating tuning fork is included among examples of oscillatory motion mainly because its prongs
ⓐ. move to and fro about their normal positions
ⓑ. move in one direction with increasing speed
ⓒ. rotate steadily about the handle of the fork
ⓓ. remain permanently bent after being struck
Correct Answer: move to and fro about their normal positions
Explanation: The prongs of a tuning fork vibrate about their normal positions after the fork is struck. This is an oscillatory motion because the prongs repeatedly move on opposite sides of an equilibrium position. The motion is called vibratory because the oscillations are rapid. Steady rotation or one-way motion would not describe the back-and-forth nature of the prong motion. The normal position is important because displacement is measured from that reference position.
5. For an oscillator, the symbol \(x\) most commonly represents
ⓐ. the total distance covered in many oscillations
ⓑ. the time taken for one complete oscillation
ⓒ. the mass of the oscillating body
ⓓ. instantaneous displacement from mean position
Correct Answer: instantaneous displacement from mean position
Explanation: In oscillations, \(x\) usually denotes the displacement of the body from its mean position at a particular instant. It can be positive on one side of the mean position and negative on the other side, depending on the chosen sign convention. The total distance travelled is different from \(x\), because distance keeps adding while displacement depends on position relative to the mean point. Mass is usually denoted by \(m\), and the time period is denoted by \(T\). The sign of \(x\) helps describe which side of the mean position the body is on.
6. Study the table and identify the row in which the symbol and unit are not properly paired.
| Row | Quantity | Usual symbol | SI unit |
|---|
| P | Displacement from mean position | \(x\) | \(\text{m}\) |
| Q | Amplitude | \(A\) | \(\text{m}\) |
| R | Time period | \(T\) | \(\text{s}\) |
| S | Frequency | \(f\) | \(\text{rad s}^{-1}\) |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: Displacement \(x\) and amplitude \(A\) are lengths, so their SI unit is \(\text{m}\). Time period \(T\) is the time for one complete oscillation, so its unit is \(\text{s}\). Frequency \(f\) means the number of oscillations per unit time, and its SI unit is \(\text{Hz}\) or \(\text{s}^{-1}\). The unit \(\text{rad s}^{-1}\) is used for angular frequency \(\omega\), not ordinary frequency \(f\). The factor \(2\pi\) separates \(\omega\) from \(f\), so their units should not be interchanged casually.
7. A block attached to a horizontal spring is pulled slightly and released on a smooth surface. The motion is described as oscillatory because the block
ⓐ. has a large mass compared with the spring
ⓑ. moves only when an external force keeps pushing it
ⓒ. moves back and forth about its equilibrium position
ⓓ. covers equal distances in every equal interval of time
Correct Answer: moves back and forth about its equilibrium position
Explanation: A spring-block system can oscillate because the spring provides a restoring force when the block is displaced from equilibrium. After release, the block moves toward the equilibrium position and may continue to the other side due to inertia. This repeated back-and-forth motion is the key feature of oscillation. Equal distances in equal intervals would suggest uniform motion, not the usual motion of an oscillator. The equilibrium position is the reference point about which the displacement changes sign.
8. The amplitude \(A\) of an oscillator is
ⓐ. maximum displacement from the mean position
ⓑ. the total distance travelled in one complete oscillation
ⓒ. the time required to go from one extreme to the other
ⓓ. the number of oscillations completed per second
Correct Answer: maximum displacement from the mean position
Explanation: Amplitude \(A\) is the greatest distance of the oscillator from its mean position on either side. It is a length, so its SI unit is \(\text{m}\). The total distance travelled in one complete oscillation may be related to amplitude in special cases, but it is not the definition of amplitude. Time period and frequency describe timing, while amplitude describes the maximum displacement. Amplitude is measured from the mean position, not from one extreme position to the other.
9. During one complete oscillation, the time taken is called the
ⓐ. frequency
ⓑ. phase constant
ⓒ. amplitude
ⓓ. time period
Correct Answer: time period
Explanation: The time period \(T\) is the time taken by an oscillating body to complete one full oscillation. It is measured in seconds, written as \(\text{s}\). Frequency \(f\) counts how many oscillations occur in one second, so it is linked to \(T\) but is not the same quantity. Amplitude \(A\) gives the maximum displacement, not the duration of the motion. A complete oscillation means the body returns to the same state of motion, not merely to the same position once.
10. An oscillator completes \(12\) oscillations in \(3.0\,\text{s}\). Its frequency is
ⓐ. \(3.0\,\text{Hz}\)
ⓑ. \(9.0\,\text{Hz}\)
ⓒ. \(4.0\,\text{Hz}\)
ⓓ. \(36\,\text{Hz}\)
Correct Answer: \(4.0\,\text{Hz}\)
Explanation: \( \textbf{Given data:} \) Number of oscillations \(N=12\), time taken \(t=3.0\,\text{s}\).
\( \textbf{Required quantity:} \) Frequency \(f\).
Frequency means number of complete oscillations per unit time.
The useful relation is
\[
f=\frac{N}{t}
\]
Substitute the given values:
\[
f=\frac{12}{3.0\,\text{s}}
\]
Calculate the quotient:
\[
f=4.0\,\text{s}^{-1}
\]
Since \(\text{s}^{-1}\) for oscillations is written as \(\text{Hz}\), the result is \(4.0\,\text{Hz}\).
The option \(36\,\text{Hz}\) comes from multiplying \(12\) and \(3.0\), but frequency requires division by the total time.
\( \textbf{Final answer:} \) \(f=4.0\,\text{Hz}\).
11. The unit \(\text{Hz}\) is used for the quantity that tells
ⓐ. maximum displacement from mean position
ⓑ. displacement at a particular instant
ⓒ. time taken for one oscillation
ⓓ. number of oscillations per second
Correct Answer: number of oscillations per second
Explanation: The unit \(\text{Hz}\) is used for frequency \(f\). A frequency of \(1\,\text{Hz}\) means one complete oscillation per second. Maximum displacement is amplitude \(A\), and its SI unit is \(\text{m}\). The time taken for one oscillation is time period \(T\), and its unit is \(\text{s}\). Frequency and time period are reciprocal quantities, so their meanings should not be swapped.
12. For an oscillator with time period \(T=0.50\,\text{s}\), the frequency is
ⓐ. \(2.0\,\text{Hz}\)
ⓑ. \(0.50\,\text{Hz}\)
ⓒ. \(1.0\,\text{Hz}\)
ⓓ. \(4.0\,\text{Hz}\)
Correct Answer: \(2.0\,\text{Hz}\)
Explanation: \( \textbf{Given:} \) Time period \(T=0.50\,\text{s}\).
\( \textbf{Required:} \) Frequency \(f\).
For periodic or oscillatory motion, frequency and time period are related by
\[
f=\frac{1}{T}
\]
This relation applies because \(T\) is the time for one oscillation, while \(f\) is the number of oscillations per second.
Substitute \(T=0.50\,\text{s}\):
\[
f=\frac{1}{0.50\,\text{s}}
\]
\[
f=2.0\,\text{s}^{-1}
\]
For oscillations, \(\text{s}^{-1}\) is written as \(\text{Hz}\).
The value \(0.50\,\text{Hz}\) would repeat the time period as if it were frequency, which reverses the reciprocal relation.
\( \textbf{Final answer:} \) \(f=2.0\,\text{Hz}\).
13. A note about an oscillator says: “The body is at the extreme position when \(|x|=A\).” This statement means that
ⓐ. the body is at the mean position
ⓑ. the body has completed exactly one oscillation
ⓒ. the displacement has maximum magnitude
ⓓ. the frequency has become equal to the amplitude
Correct Answer: the displacement has maximum magnitude
Explanation: The symbol \(x\) represents instantaneous displacement from the mean position. The symbol \(A\) represents amplitude, which is the maximum magnitude of \(x\). Therefore, \(|x|=A\) means the body is at one of its extreme positions. The absolute value is used because the extreme can be on the \(+\) side or the \(-\) side of the mean position. Frequency is a time-related quantity, so it cannot become equal to a length like amplitude.
14. In an oscillating system, the restoring tendency is directed
ⓐ. away from the mean position on both sides
ⓑ. only when the body is exactly at an extreme position
ⓒ. only along the positive side of the chosen axis
ⓓ. toward the mean position when the body is displaced
Correct Answer: toward the mean position when the body is displaced
Explanation: A restoring tendency is the tendency that tries to bring the displaced body back toward its equilibrium or mean position. If the body is displaced to the positive side, the restoring effect is toward the negative direction, and if it is displaced to the negative side, it is toward the positive direction. This direction is why the motion can reverse again and again. The restoring tendency need not always have the same direction in space; it depends on which side of the mean position the body occupies. At the mean position, the displacement from equilibrium is zero, so the restoring effect is not described as pulling it farther away.
15. Match the basic oscillation symbols with their usual meanings.
| Column I | Column II |
|---|
| P. \(A\) | 1. Time for one complete oscillation |
| Q. \(T\) | 2. Number of oscillations per second |
| R. \(f\) | 3. Maximum displacement from mean position |
| S. \(\omega\) | 4. Angular frequency |
ⓐ. P-3, Q-1, R-2, S-4
ⓑ. P-1, Q-3, R-2, S-4
ⓒ. P-3, Q-2, R-1, S-4
ⓓ. P-4, Q-1, R-2, S-3
Correct Answer: P-3, Q-1, R-2, S-4
Explanation: Amplitude \(A\) is the maximum displacement from the mean position. Time period \(T\) is the time required for one complete oscillation. Frequency \(f\) is the number of complete oscillations per second. Angular frequency \(\omega\) is related to frequency by \(\omega=2\pi f\). The distinction between \(f\) and \(\omega\) matters because \(\omega\) measures angular phase change per unit time, while \(f\) counts cycles per unit time.
16. Use the graph description below.
The displacement \(x\) of an oscillating body is plotted against time \(t\). The curve repeats smoothly above and below the mean line \(x=0\), and the highest positive value of \(x\) is \(+A\).
The value \(+A\) on this graph represents
ⓐ. one full time period
ⓑ. the frequency of the motion
ⓒ. the angular frequency of the motion
ⓓ. the positive extreme displacement
Correct Answer: the positive extreme displacement
Explanation: On an \(x-t\) graph, the vertical coordinate gives displacement \(x\) from the mean position. The line \(x=0\) represents the mean position. The highest positive value \(+A\) is the extreme displacement on the positive side, so its magnitude is the amplitude. Time period would be read along the time axis as the interval after which the graph repeats. Frequency and angular frequency are obtained from timing information, not from the vertical height alone.
17. Use the arrangement described below. A pendulum bob moves along an arc through three marked positions: left extreme \(L\), lowest point \(M\), and right extreme \(R\). For small oscillations, the position usually treated as the mean position is
ⓐ. \(L\), because the bob pauses there momentarily
ⓑ. \(R\), because the bob pauses there momentarily
ⓒ. \(M\), because it is the equilibrium position
ⓓ. any point on the arc, because all positions repeat
Correct Answer: \(M\), because it is the equilibrium position
Explanation: In a simple pendulum, the lowest point \(M\) is the equilibrium or mean position for small oscillations. The bob moves to and fro about this point as it swings between the two extremes. At \(L\) and \(R\), the bob is farthest from the mean position, so those are extreme positions rather than the mean position. The bob may momentarily stop at an extreme, but stopping for an instant does not make that point the equilibrium position. The mean position is chosen from force balance and symmetry, not from where the bob’s speed happens to be zero.
18. The relation connecting angular frequency \(\omega\) and ordinary frequency \(f\) is best completed as \(\omega=\underline{\hspace{1.5cm}}\).
ⓐ. \(2\pi f\)
ⓑ. \(\frac{f}{2\pi}\)
ⓒ. \(fT\)
ⓓ. \(\frac{T}{2\pi}\)
Correct Answer: \(2\pi f\)
Explanation: \( \textbf{Known idea:} \) Ordinary frequency \(f\) counts complete cycles per second.
Angular frequency \(\omega\) measures angular phase change per second.
One complete cycle corresponds to an angular change of \(2\pi\,\text{rad}\).
So, angular change per second equals \(2\pi\) times the number of cycles per second.
The relation is
\[
\omega=2\pi f
\]
The unit of \(f\) is \(\text{Hz}\), or \(\text{s}^{-1}\).
The unit of \(\omega\) is commonly written as \(\text{rad s}^{-1}\).
Using \(\frac{f}{2\pi}\) would reduce the value instead of converting cycles per second into radians per second.
The factor \(2\pi\) appears because each complete oscillation corresponds to one full phase cycle.
\( \textbf{Final answer:} \) \(\omega=2\pi f\).
19. A body passes through its mean position during an oscillation. At that instant, its displacement \(x\) from the mean position is
ⓐ. \(A\)
ⓑ. \(-A\)
ⓒ. \(0\)
ⓓ. \(2A\)
Correct Answer: \(0\)
Explanation: The displacement \(x\) in oscillatory motion is measured from the mean or equilibrium position. When the body is exactly at the mean position, its position coincides with the reference point from which \(x\) is measured. Therefore, the instantaneous displacement is \(0\), even though the body may be moving. The values \(+A\) and \(-A\) correspond to extreme positions on the two sides of the mean position. Displacement should not be confused with speed, because a body can have zero displacement from the mean position and still be in motion.
20. A motion has frequency \(5\,\text{Hz}\). The number of complete oscillations made in \(20\,\text{s}\) is
ⓐ. \(4\)
ⓑ. \(25\)
ⓒ. \(100\)
ⓓ. \(0.25\)
Correct Answer: \(100\)
Explanation: \( \textbf{Given data:} \) Frequency \(f=5\,\text{Hz}\) and time \(t=20\,\text{s}\).
Frequency \(5\,\text{Hz}\) means \(5\) complete oscillations occur in \(1\,\text{s}\).
The required number of oscillations is \(N\).
Use the relation
\[
N=ft
\]
Substitute the values:
\[
N=(5\,\text{s}^{-1})(20\,\text{s})
\]
The unit \(\text{s}^{-1}\) cancels with \(\text{s}\).
\[
N=100
\]
The answer is a pure number because it counts complete oscillations.
The option \(4\) would come from using \(\frac{20}{5}\), which gives time per oscillation only if the given \(5\) were a time period.
\( \textbf{Final answer:} \) \(100\) complete oscillations.
