101. The differential equation \(\frac{d^2x}{dt^2}+\omega^2x=0\) expresses SHM because it means
ⓐ. velocity is always zero
ⓑ. acceleration is independent of displacement
ⓒ. displacement is constant with time
ⓓ. acceleration is linear and restoring
Correct Answer: acceleration is linear and restoring
Explanation: The second derivative \(\frac{d^2x}{dt^2}\) represents acceleration \(a\). From \(\frac{d^2x}{dt^2}+\omega^2x=0\), we get \(\frac{d^2x}{dt^2}=-\omega^2x\). This means acceleration is proportional to displacement and has the opposite sign. That is the defining condition of SHM. The equation does not say the body is always at rest; instead, it describes how acceleration changes with position.
102. A proposed equation of motion is \(\frac{d^2x}{dt^2}-9x=0\), where \(x\) is measured from the mean position. This does not represent SHM because it gives
ⓐ. \(a=-9x\)
ⓑ. \(x=0\) at all times
ⓒ. \(a=0\)
ⓓ. \(a=+9x\)
Correct Answer: \(a=+9x\)
Explanation: \( \textbf{Given equation:} \)
\[
\frac{d^2x}{dt^2}-9x=0
\]
Move the displacement term to the other side:
\[
\frac{d^2x}{dt^2}=9x
\]
Since acceleration is
\[
a=\frac{d^2x}{dt^2}
\]
the equation gives
\[
a=+9x
\]
For SHM, the required form is \(a=-\omega^2x\).
Here, acceleration has the same sign as displacement, so it is not a restoring acceleration toward the mean position.
\( \textbf{Final answer:} \) The equation gives \(a=+9x\), not the SHM condition.
103. Match the mathematical statement with its physical meaning.
| Column I | Column II |
|---|
| P. \(x=0\) | 1. Mean position |
| Q. \(a=-\omega^2x\) | 2. Restoring acceleration condition |
| R. \(\frac{d^2x}{dt^2}+\omega^2x=0\) | 3. Differential equation of SHM |
| S. \(|x|=A\) | 4. Extreme position |
ⓐ. P-1, Q-2, R-3, S-4
ⓑ. P-4, Q-3, R-2, S-1
ⓒ. P-1, Q-3, R-4, S-2
ⓓ. P-2, Q-1, R-3, S-4
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: The value \(x=0\) means the body is at the mean position because displacement is measured from that point. The relation \(a=-\omega^2x\) states the restoring acceleration condition for SHM. The equation \(\frac{d^2x}{dt^2}+\omega^2x=0\) is the differential equation form of the same condition. The condition \(|x|=A\) means the displacement has maximum magnitude, so the body is at an extreme position. These four statements describe different but connected parts of the SHM description.
104. A solution \(x=A\sin(\omega t+\phi)\) satisfies the SHM differential equation because its second derivative is
ⓐ. equal to \(+\omega^2x\)
ⓑ. equal to \(-\omega^2x\)
ⓒ. independent of \(x\)
ⓓ. always equal to \(A\omega\)
Correct Answer: equal to \(-\omega^2x\)
Explanation: \( \textbf{Given solution:} \)
\[
x=A\sin(\omega t+\phi)
\]
Differentiate once:
\[
\frac{dx}{dt}=A\omega\cos(\omega t+\phi)
\]
Differentiate again:
\[
\frac{d^2x}{dt^2}=-A\omega^2\sin(\omega t+\phi)
\]
Since
\[
x=A\sin(\omega t+\phi)
\]
we get
\[
\frac{d^2x}{dt^2}=-\omega^2x
\]
Substitute this into
\[
\frac{d^2x}{dt^2}+\omega^2x=0
\]
The two terms cancel, so the sine form satisfies the SHM equation.
\( \textbf{Final answer:} \) \(\frac{d^2x}{dt^2}=-\omega^2x\).
105. A body of mass \(0.50\,\text{kg}\) moves under a force \(F=-8x\), where \(F\) is in \(\text{N}\) and \(x\) is in \(\text{m}\). Its angular frequency is
ⓐ. \(2\,\text{rad s}^{-1}\)
ⓑ. \(4\,\text{rad s}^{-1}\)
ⓒ. \(8\,\text{rad s}^{-1}\)
ⓓ. \(16\,\text{rad s}^{-1}\)
Correct Answer: \(4\,\text{rad s}^{-1}\)
Explanation: \( \textbf{Given:} \) \(m=0.50\,\text{kg}\) and \(F=-8x\).
Compare \(F=-8x\) with the restoring force form:
\[
F=-kx
\]
So,
\[
k=8\,\text{N m}^{-1}
\]
Newton's second law gives
\[
F=ma
\]
Thus,
\[
ma=-kx
\]
\[
a=-\frac{k}{m}x
\]
Compare with the SHM relation:
\[
a=-\omega^2x
\]
So,
\[
\omega^2=\frac{k}{m}
\]
Substitute the values:
\[
\omega^2=\frac{8}{0.50}=16
\]
\[
\omega=4\,\text{rad s}^{-1}
\]
The mass must be included because the same restoring force constant gives different acceleration for different masses.
\( \textbf{Final answer:} \) \(\omega=4\,\text{rad s}^{-1}\).
106. A motion is said to be simple harmonic only if the restoring influence is linear for the displacement range used. This condition is needed because
ⓐ. SHM needs linear restoring acceleration
ⓑ. SHM requires speed to remain constant throughout the path
ⓒ. SHM requires displacement to increase forever with time
ⓓ. SHM requires the force to be independent of displacement
Correct Answer: SHM needs linear restoring acceleration
Explanation: Simple harmonic motion is based on a linear restoring condition. In acceleration form, this is \(a=-\omega^2x\), and in force form for a fixed mass, it corresponds to \(F\propto -x\). The proportionality must be linear so that doubling the displacement doubles the magnitude of restoring acceleration. If the force is nonlinear, the motion may still be oscillatory, but it is not simple harmonic in the strict ideal form. Constant speed is not required in SHM; in fact, speed changes continuously during the motion.
107. For a linear oscillator obeying \(F=-kx\), the constant \(k\) represents
ⓐ. the time taken for one complete oscillation
ⓑ. the restoring force per unit displacement
ⓒ. the maximum displacement from the mean position
ⓓ. the phase change in one second
Correct Answer: the restoring force per unit displacement
Explanation: In the relation \(F=-kx\), the magnitude of restoring force is \(kx\). For a given displacement \(x\), a larger \(k\) gives a larger restoring force. Thus, \(k\) measures how strongly the system resists displacement from equilibrium. Its unit is \(\text{N m}^{-1}\), because force is measured in \(\text{N}\) and displacement in \(\text{m}\). The negative sign is not part of the size of \(k\); it shows that the force acts opposite to displacement.
108. The SI unit of the force constant \(k\) in \(F=-kx\) is
ⓐ. \(\text{N m}^{-1}\)
ⓑ. \(\text{N m}\)
ⓒ. \(\text{kg m s}^{-1}\)
ⓓ. \(\text{m s}^{-2}\)
Correct Answer: \(\text{N m}^{-1}\)
Explanation: From the restoring force relation,
\[
F=-kx
\]
the magnitude relation is
\[
k=\frac{|F|}{|x|}
\]
Force \(F\) is measured in \(\text{N}\).
Displacement \(x\) is measured in \(\text{m}\).
Therefore,
\[
[k]=\frac{\text{N}}{\text{m}}
\]
\[
[k]=\text{N m}^{-1}
\]
Since \(1\,\text{N}=1\,\text{kg m s}^{-2}\), this is also equivalent to \(\text{kg s}^{-2}\).
The unit \(\text{N m}\) belongs to work or energy type quantities, not to force per unit displacement.
\( \textbf{Final answer:} \) The unit of \(k\) is \(\text{N m}^{-1}\).
109. Study the table for a spring oscillator in which displacement to the right is positive.
| Row | Displacement | Restoring force from \(F=-kx\) |
|---|
| P | \(x\gt0\) | Positive |
| Q | \(x\gt0\) | Negative |
| R | \(x\lt0\) | Negative |
| S | \(x=0\) | Maximum positive |
The row that correctly gives the force direction is
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row Q
Explanation: The restoring force relation is \(F=-kx\), where \(k\) is positive. If \(x\gt0\), the displacement is to the positive side, so the force is negative. A negative force acts toward the left if the right side is chosen positive. This is the restoring direction because it points back toward the mean position. If \(x=0\), the spring force is zero, not maximum, because the spring is at its equilibrium length for this displacement coordinate.
110. A spring is stretched by \(0.040\,\text{m}\) and exerts a restoring force of magnitude \(2.0\,\text{N}\). The force constant of the spring is
ⓐ. \(0.080\,\text{N m}^{-1}\)
ⓑ. \(8.0\,\text{N m}^{-1}\)
ⓒ. \(50\,\text{N m}^{-1}\)
ⓓ. \(80\,\text{N m}^{-1}\)
Correct Answer: \(50\,\text{N m}^{-1}\)
Explanation: \( \textbf{Given:} \) Restoring force magnitude \(|F|=2.0\,\text{N}\), displacement magnitude \(|x|=0.040\,\text{m}\).
For a linear spring,
\[
|F|=k|x|
\]
The force constant is
\[
k=\frac{|F|}{|x|}
\]
Substitute the values:
\[
k=\frac{2.0\,\text{N}}{0.040\,\text{m}}
\]
\[
k=50\,\text{N m}^{-1}
\]
The sign in \(F=-kx\) gives direction, but the force constant itself is reported as a positive stiffness value.
The option \(0.080\,\text{N m}^{-1}\) comes from multiplying force and displacement instead of dividing.
\( \textbf{Final answer:} \) \(k=50\,\text{N m}^{-1}\).
111. A block of mass \(0.25\,\text{kg}\) is attached to a spring of force constant \(100\,\text{N m}^{-1}\) on a smooth horizontal surface. The angular frequency of small oscillations is
ⓐ. \(5\,\text{rad s}^{-1}\)
ⓑ. \(10\,\text{rad s}^{-1}\)
ⓒ. \(20\,\text{rad s}^{-1}\)
ⓓ. \(400\,\text{rad s}^{-1}\)
Correct Answer: \(20\,\text{rad s}^{-1}\)
Explanation: \( \textbf{Given:} \) \(m=0.25\,\text{kg}\), \(k=100\,\text{N m}^{-1}\).
For a horizontal spring-mass oscillator,
\[
F=-kx
\]
Using Newton's second law,
\[
ma=-kx
\]
So,
\[
a=-\frac{k}{m}x
\]
Compare with the SHM relation:
\[
a=-\omega^2x
\]
Therefore,
\[
\omega^2=\frac{k}{m}
\]
Substitute the values:
\[
\omega^2=\frac{100}{0.25}=400
\]
\[
\omega=20\,\text{rad s}^{-1}
\]
The square root step is essential because \(\frac{k}{m}\) gives \(\omega^2\), not \(\omega\).
\( \textbf{Final answer:} \) \(\omega=20\,\text{rad s}^{-1}\).
112. Two spring-mass systems have the same mass, but spring \(P\) has a larger force constant than spring \(Q\). If both behave ideally, spring \(P\) gives
ⓐ. a smaller angular frequency because the restoring force is weaker
ⓑ. a larger angular frequency because \(k\) is larger
ⓒ. the same angular frequency because amplitude alone controls \(\omega\)
ⓓ. zero angular frequency because a stiff spring cannot oscillate
Correct Answer: a larger angular frequency because \(k\) is larger
Explanation: For an ideal spring-mass system, the angular frequency is \(\omega=\sqrt{\frac{k}{m}}\). If the mass \(m\) is the same, increasing \(k\) increases \(\frac{k}{m}\). Therefore, \(\omega\) becomes larger. A larger \(k\) means the spring gives a stronger restoring force for the same displacement. In the ideal formula, amplitude does not decide \(\omega\); the system parameters \(k\) and \(m\) do.
113. Read the situation below and answer the question.
A block attached to a spring is displaced \(5\,\text{cm}\) to the left of its equilibrium position. The positive direction is chosen to the right. The spring obeys \(F=-kx\).
The signs of displacement and restoring force are respectively
ⓐ. negative and positive
ⓑ. positive and negative
ⓒ. negative and negative
ⓓ. positive and positive
Correct Answer: negative and positive
Explanation: Since the block is to the left of equilibrium and the right direction is positive, its displacement is negative. The force relation is \(F=-kx\), where \(k\) is positive. A negative value of \(x\) makes \(F\) positive. A positive force points to the right, which is toward the equilibrium position from the left side. The signs are not arbitrary; they encode the restoring nature of the spring force.
114. Assertion: Increasing the force constant \(k\) of an ideal spring-mass system increases its angular frequency if the mass is unchanged.
Reason: For an ideal spring-mass system, \(\omega=\sqrt{\frac{k}{m}}\).
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The angular frequency of an ideal spring-mass oscillator is \(\omega=\sqrt{\frac{k}{m}}\). If \(m\) is constant, increasing \(k\) increases the ratio \(\frac{k}{m}\). Taking the square root still gives a larger value of \(\omega\). Physically, a stiffer spring gives a stronger restoring force for the same displacement, so the system returns more quickly. The Reason gives the exact mathematical dependence needed to explain the Assertion.
115. In SHM, the magnitude of acceleration is maximum when the body is
ⓐ. at the mean position
ⓑ. halfway between mean position and extreme only
ⓒ. at either extreme position
ⓓ. moving with maximum speed through the mean position
Correct Answer: at either extreme position
Explanation: In SHM, acceleration is related to displacement by \(a=-\omega^2x\). The magnitude is \(|a|=\omega^2|x|\). This magnitude is largest when \(|x|\) is largest. The largest possible value of \(|x|\) is the amplitude \(A\), which occurs at the extreme positions. At the mean position, \(x=0\), so acceleration is zero even though the speed may be maximum there.
116. For a body in SHM with positive direction to the right, the body is at the right extreme position. Its acceleration is
ⓐ. zero
ⓑ. toward the right with maximum magnitude
ⓒ. toward the left with maximum magnitude
ⓓ. toward the right with minimum nonzero magnitude
Correct Answer: toward the left with maximum magnitude
Explanation: At the right extreme position, displacement \(x\) is positive and has maximum magnitude \(A\). In SHM, \(a=-\omega^2x\), so a positive displacement gives a negative acceleration. Negative acceleration means acceleration toward the left when the right side is chosen positive. The magnitude is maximum because \(|x|=A\) at the extreme. The acceleration is not zero at the extreme; the velocity is zero there in ideal SHM.
117. A body in SHM has amplitude \(0.050\,\text{m}\) and angular frequency \(6\,\text{rad s}^{-1}\). The maximum magnitude of its acceleration is
ⓐ. \(0.30\,\text{m s}^{-2}\)
ⓑ. \(1.8\,\text{m s}^{-2}\)
ⓒ. \(3.6\,\text{m s}^{-2}\)
ⓓ. \(36\,\text{m s}^{-2}\)
Correct Answer: \(1.8\,\text{m s}^{-2}\)
Explanation: \( \textbf{Given:} \) Amplitude \(A=0.050\,\text{m}\), angular frequency \(\omega=6\,\text{rad s}^{-1}\).
The acceleration in SHM is
\[
a=-\omega^2x
\]
The magnitude is
\[
|a|=\omega^2|x|
\]
Maximum acceleration occurs at the extreme position where \(|x|=A\).
So,
\[
a_{\max}=\omega^2A
\]
Substitute the values:
\[
a_{\max}=(6)^2(0.050)
\]
\[
a_{\max}=36\times0.050
\]
\[
a_{\max}=1.8\,\text{m s}^{-2}
\]
The amplitude must be used in metres because acceleration is required in \(\text{m s}^{-2}\).
\( \textbf{Final answer:} \) \(a_{\max}=1.8\,\text{m s}^{-2}\).
118. Use the graph description below.
A graph of acceleration \(a\) versus displacement \(x\) for an oscillator is a straight line through the origin. Its slope is \(-25\,\text{s}^{-2}\).
The angular frequency of the oscillator is
ⓐ. \(25\,\text{rad s}^{-1}\)
ⓑ. \(12.5\,\text{rad s}^{-1}\)
ⓒ. \(5\,\text{rad s}^{-1}\)
ⓓ. \(\frac{1}{5}\,\text{rad s}^{-1}\)
Correct Answer: \(5\,\text{rad s}^{-1}\)
Explanation: \( \textbf{Graph relation:} \) For SHM,
\[
a=-\omega^2x
\]
In an \(a\)-versus-\(x\) graph, the slope is the coefficient of \(x\):
\[
\text{slope}=-\omega^2
\]
Given:
\[
\text{slope}=-25\,\text{s}^{-2}
\]
Therefore,
\[
-\omega^2=-25\,\text{s}^{-2}
\]
\[
\omega^2=25\,\text{s}^{-2}
\]
Taking the positive square root,
\[
\omega=5\,\text{rad s}^{-1}
\]
The negative sign of the slope gives the restoring direction, while the square root gives the angular frequency.
\( \textbf{Final answer:} \) \(\omega=5\,\text{rad s}^{-1}\).
119. A student record says: “At the mean position in SHM, both acceleration and velocity must be zero.” The part that needs correction is that
ⓐ. acceleration is not zero at the mean position
ⓑ. velocity need not be zero at the mean position
ⓒ. displacement is maximum at the mean position
ⓓ. acceleration is always maximum at the mean position
Correct Answer: velocity need not be zero at the mean position
Explanation: At the mean position of SHM, displacement \(x\) is zero. From \(a=-\omega^2x\), the acceleration is also zero there. However, the body generally passes through the mean position with maximum speed in ideal SHM. Zero acceleration does not mean zero velocity; it only means the velocity is not changing instantaneously at that point. The velocity becomes zero at the extreme positions, where displacement and acceleration magnitude are maximum.
120. A body in SHM has maximum acceleration magnitude \(12\,\text{m s}^{-2}\) and amplitude \(0.030\,\text{m}\). Its angular frequency is
ⓐ. \(10\,\text{rad s}^{-1}\)
ⓑ. \(20\,\text{rad s}^{-1}\)
ⓒ. \(40\,\text{rad s}^{-1}\)
ⓓ. \(400\,\text{rad s}^{-1}\)
Correct Answer: \(20\,\text{rad s}^{-1}\)
Explanation: \( \textbf{Given:} \) \(a_{\max}=12\,\text{m s}^{-2}\), \(A=0.030\,\text{m}\).
For SHM, maximum acceleration is
\[
a_{\max}=\omega^2A
\]
Rearrange for \(\omega^2\):
\[
\omega^2=\frac{a_{\max}}{A}
\]
Substitute the values:
\[
\omega^2=\frac{12}{0.030}
\]
\[
\omega^2=400\,\text{s}^{-2}
\]
Now take the positive square root:
\[
\omega=20\,\text{rad s}^{-1}
\]
The value \(400\) is \(\omega^2\), so choosing \(400\,\text{rad s}^{-1}\) skips the square-root step.
\( \textbf{Final answer:} \) \(\omega=20\,\text{rad s}^{-1}\).
