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101. Hooke’s law states that the restoring force in a spring is:
ⓐ. Directly proportional to the mass of the object
ⓑ. Directly proportional to displacement from equilibrium
ⓒ. Inversely proportional to displacement
ⓓ. Independent of displacement
Correct Answer: Directly proportional to displacement from equilibrium
Explanation: Hooke’s law is written as $F = -kx$, where $k$ is the spring constant and $x$ is the displacement. The negative sign shows that the force is opposite to the displacement.
102. The spring constant $k$ is defined as:
ⓐ. Force per unit velocity
ⓑ. Force per unit displacement
ⓒ. Displacement per unit mass
ⓓ. Energy per unit time
Correct Answer: Force per unit displacement
Explanation: The spring constant measures stiffness: $k = \frac{F}{x}$. A larger $k$ means a stiffer spring requiring more force for the same displacement.
103. A spring stretches by $0.05 \, \text{m}$ when a force of $10 \, \text{N}$ is applied. What is the spring constant?
ⓐ. $100 \, \text{N/m}$
ⓑ. $200 \, \text{N/m}$
ⓒ. $150 \, \text{N/m}$
ⓓ. $50 \, \text{N/m}$
Correct Answer: $200 \, \text{N/m}$
Explanation: $k = \frac{F}{x} = \frac{10}{0.05} = 200 \, \text{N/m}$. This means the spring is relatively stiff.
104. Which physical quantity has SI unit $\text{N/m}$?
ⓐ. Displacement
ⓑ. Frequency
ⓒ. Spring constant
ⓓ. Energy
Correct Answer: Spring constant
Explanation: The spring constant $k$ has units $\text{N/m}$, since it is defined as force per unit displacement.
105. A spring of constant $k = 300 \, \text{N/m}$ is compressed by $0.1 \, \text{m}$. What is the restoring force?
ⓐ. $10 \, \text{N}$
ⓑ. $20 \, \text{N}$
ⓒ. $30 \, \text{N}$
ⓓ. $40 \, \text{N}$
Correct Answer: $30 \, \text{N}$
Explanation: From Hooke’s law, $F = kx = 300 \times 0.1 = 30 \, \text{N}$. The negative sign shows force acts opposite displacement.
106. The potential energy stored in a stretched spring is given by:
ⓐ. $U = \frac{1}{2}kx^2$
ⓑ. $U = kx$
ⓒ. $U = Fx$
ⓓ. $U = \frac{1}{2}mv^2$
Correct Answer: $U = \frac{1}{2}kx^2$
Explanation: Work done in stretching the spring is stored as potential energy. $U = \int_0^x F dx = \int_0^x kx dx = \frac{1}{2}kx^2$.
107. A spring with $k = 400 \, \text{N/m}$ is compressed by $0.05 \, \text{m}$. What is the potential energy stored?
ⓐ. $0.25 \, \text{J}$
ⓑ. $0.5 \, \text{J}$
ⓒ. $1.0 \, \text{J}$
ⓓ. $2.0 \, \text{J}$
Correct Answer: $0.5 \, \text{J}$
Explanation: $U = \frac{1}{2}kx^2 = \frac{1}{2} \times 400 \times (0.05)^2 = 0.5 \, \text{J}$.
108. If the mass $m$ is attached to a spring with constant $k$, the angular frequency of oscillations is:
ⓐ. $\omega = \frac{k}{m}$
ⓑ. $\omega = \sqrt{\frac{k}{m}}$
ⓒ. $\omega = \sqrt{\frac{m}{k}}$
ⓓ. $\omega = \frac{m}{k}$
Correct Answer: $\omega = \sqrt{\frac{k}{m}}$
Explanation: From Newton’s law and Hooke’s law, $m\ddot{x} + kx = 0$. The standard SHM equation gives $\omega = \sqrt{\frac{k}{m}}$.
109. A mass of $0.5 \, \text{kg}$ is attached to a spring of constant $200 \, \text{N/m}$. What is the time period of oscillation?
ⓐ. $0.22 \, \text{s}$
ⓑ. $0.28 \, \text{s}$
ⓒ. $0.31 \, \text{s}$
ⓓ. $0.62 \, \text{s}$
Correct Answer: $0.31 \, \text{s}$
Explanation: Time period $T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.5}{200}} = 2\pi \sqrt{0.0025} = 2\pi \times 0.05 = 0.314 \, \text{s}$. Approx answer: $0.31 \, \text{s}$.
110. Two springs of constants $k_1 = 200 \, \text{N/m}$ and $k_2 = 300 \, \text{N/m}$ are connected in parallel. What is the equivalent spring constant?
ⓐ. $100 \, \text{N/m}$
ⓑ. $200 \, \text{N/m}$
ⓒ. $300 \, \text{N/m}$
ⓓ. $500 \, \text{N/m}$
Correct Answer: $500 \, \text{N/m}$
Explanation: For parallel springs, $k_{\text{eq}} = k_1 + k_2 = 200 + 300 = 500 \, \text{N/m}$. For series, the reciprocal rule applies.
111. What is the equation of motion for a mass $m$ attached to a spring of constant $k$?
ⓐ. $m\ddot{x} = kx$
ⓑ. $m\ddot{x} + kx = 0$
ⓒ. $\ddot{x} = 0$
ⓓ. $m\ddot{x} – kx = 0$
Correct Answer: $m\ddot{x} + kx = 0$
Explanation: The restoring force is $F = -kx$. By Newton’s law, $F = m\ddot{x}$. Hence, $m\ddot{x} + kx = 0$, which is the standard SHM equation for a spring-mass system.
112. A $0.2 \, \text{kg}$ mass attached to a spring stretches by $0.04 \, \text{m}$ under gravity. What is the spring constant?
ⓐ. $25 \, \text{N/m}$
ⓑ. $50 \, \text{N/m}$
ⓒ. $75 \, \text{N/m}$
ⓓ. $100 \, \text{N/m}$
Correct Answer: $50 \, \text{N/m}$
Explanation: At equilibrium, $mg = kx$. Substituting: $(0.2 \times 9.8) = k(0.04)$. Thus $k = \frac{1.96}{0.04} = 49 \approx 50 \, \text{N/m}$.
113. The angular frequency of oscillation of a mass-spring system is:
ⓐ. $\omega = \frac{k}{m}$
ⓑ. $\omega = \sqrt{\frac{k}{m}}$
ⓒ. $\omega = \frac{m}{k}$
ⓓ. $\omega = \frac{1}{km}$
Correct Answer: $\omega = \sqrt{\frac{k}{m}}$
Explanation: The solution of $m\ddot{x} + kx = 0$ gives $\omega = \sqrt{\frac{k}{m}} $. This shows oscillation frequency depends on mass and stiffness of the spring.
114. A $1 \, \text{kg}$ block attached to a spring of constant $k = 100 \, \text{N/m}$ is displaced by $0.1 \, \text{m}$ and released. What is its maximum speed?
ⓐ. $5.0 \, \text{m/s}$
ⓑ. $4.0 \, \text{m/s}$
ⓒ. $2.0 \, \text{m/s}$
ⓓ. $1.0 \, \text{m/s}$
Correct Answer: $1.0 \, \text{m/s}$
Explanation: Maximum velocity is $v_{\max} = A\omega$. Here $A = 0.1$, $\omega = \sqrt{\tfrac{k}{m}} = \sqrt{\tfrac{100}{1}} = 10$. So, $v_{\max} = 0.1 \times 10 = 1.0 \, \text{m/s}$.
115. What is the time period of a spring-mass system with $m = 0.5 \, \text{kg}, k = 200 \, \text{N/m}$?
ⓐ. $0.22 \, \text{s}$
ⓑ. $0.31 \, \text{s}$
ⓒ. $0.44 \, \text{s}$
ⓓ. $0.62 \, \text{s}$
Correct Answer: $0.31 \, \text{s}$
Explanation: $T = 2\pi \sqrt{\tfrac{m}{k}} = 2\pi \sqrt{\tfrac{0.5}{200}} = 2\pi \sqrt{0.0025} = 2\pi (0.05) = 0.314 \, \text{s}$.
116. The total mechanical energy in SHM of a spring-mass system is:
ⓐ. $E = \tfrac{1}{2}kA^2$
ⓑ. $E = kA$
ⓒ. $E = \tfrac{1}{2}mv^2$ only
ⓓ. $E = \tfrac{1}{2}kx^2$ only
Correct Answer: $E = \tfrac{1}{2}kA^2$
Explanation: The total mechanical energy remains constant and depends only on amplitude and spring constant: $E = \tfrac{1}{2}kA^2$. Kinetic and potential energy exchange, but their sum stays constant.
117. A block of mass $0.25 \, \text{kg}$ attached to a spring of constant $100 \, \text{N/m}$ oscillates with amplitude $0.05 \, \text{m}$. What is its total energy?
ⓐ. $0.05 \, \text{J}$
ⓑ. $0.1 \, \text{J}$
ⓒ. $0.125 \, \text{J}$
ⓓ. $0.5 \, \text{J}$
Correct Answer: $0.125 \, \text{J}$
Explanation: $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2} \times 100 \times (0.05)^2 = 50 \times 0.0025 = 0.125 \, \text{J}$.
118. If displacement of a spring-mass system is $x = A \cos(\omega t)$, what is its velocity?
ⓐ. $v = A\omega \cos(\omega t)$
ⓑ. $v = -A\omega \sin(\omega t)$
ⓒ. $v = A^2\omega^2$
ⓓ. $v = 0$
Correct Answer: $v = -A\omega \sin(\omega t)$
Explanation: Differentiating displacement: $v = \frac{dx}{dt} = -A\omega \sin(\omega t)$. Velocity is maximum at mean position where $x = 0$.
119. The potential energy stored in the spring at displacement $x$ is:
ⓐ. $U = \tfrac{1}{2}kx^2$
ⓑ. $U = \tfrac{1}{2}mv^2$
ⓒ. $U = kx$
ⓓ. $U = Fx$
Correct Answer: $U = \tfrac{1}{2}kx^2$
Explanation: Potential energy increases quadratically with displacement. At extreme positions ($x = A$), potential energy is maximum and equal to total energy.
120. A spring-mass system oscillates with amplitude $0.1 \, \text{m}$, $k = 50 \, \text{N/m}$. What is the maximum potential energy?
ⓐ. $0.125 \, \text{J}$
ⓑ. $0.25 \, \text{J}$
ⓒ. $0.5 \, \text{J}$
ⓓ. $1.0 \, \text{J}$
Correct Answer: $0.25 \, \text{J}$
Explanation: Maximum potential energy = total energy = $\tfrac{1}{2}kA^2 = \tfrac{1}{2} \times 50 \times (0.1)^2 = 25 \times 0.01 = 0.25 \, \text{J}$.
121. In a mass–spring oscillator, the total mechanical energy is:
ⓐ. Constant, independent of time
ⓑ. Always zero
ⓒ. Increasing with time
ⓓ. Decreasing with time
Correct Answer: Constant, independent of time
Explanation: In ideal SHM without damping, total mechanical energy is conserved. It is shared between kinetic energy $\tfrac{1}{2}mv^2$ and potential energy $\tfrac{1}{2}kx^2$. Their sum remains constant: $E = \tfrac{1}{2}kA^2$.
122. At the mean position of oscillation, the energy of a spring-mass system is entirely:
ⓐ. Kinetic
ⓑ. Potential
ⓒ. Thermal
ⓓ. Gravitational
Correct Answer: Kinetic
Explanation: At mean position, displacement $x = 0$. Thus potential energy $U = \tfrac{1}{2}kx^2 = 0$. The total energy is present as kinetic energy, maximum at this point: $K = \tfrac{1}{2}kA^2$.
123. At the extreme position of oscillation, the energy of the spring-mass system is entirely:
ⓐ. Kinetic
ⓑ. Potential
ⓒ. Thermal
ⓓ. Rotational
Correct Answer: Potential
Explanation: At extreme positions, displacement $x = A$. Potential energy $U = \tfrac{1}{2}kA^2$ is maximum, and kinetic energy is zero because velocity vanishes at the turning point.
124. Which of the following represents kinetic energy of a mass-spring oscillator at displacement $x$?
ⓐ. $K = \tfrac{1}{2}kx^2$
ⓑ. $K = \tfrac{1}{2}k(A^2 – x^2)$
ⓒ. $K = \tfrac{1}{2}mv^2 = 0$
ⓓ. $K = kAx$
Correct Answer: $K = \tfrac{1}{2}k(A^2 – x^2)$
Explanation: The total energy is $\tfrac{1}{2}kA^2$. Potential energy is $\tfrac{1}{2}kx^2$. Hence, kinetic energy = Total – Potential = $\tfrac{1}{2}k(A^2 – x^2)$.
125. A spring-mass oscillator has $k = 200 \, \text{N/m}$, amplitude $A = 0.1 \, \text{m}$. What is its total energy?
ⓐ. $0.5 \, \text{J}$
ⓑ. $1.0 \, \text{J}$
ⓒ. $2.0 \, \text{J}$
ⓓ. $5.0 \, \text{J}$
Correct Answer: $1.0 \, \text{J}$
Explanation: $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2} \times 200 \times (0.1)^2 = 100 \times 0.01 = 1.0 \, \text{J}$.
126. For displacement $x = A \cos(\omega t)$, the potential energy of the system is:
ⓐ. $U = \tfrac{1}{2}kA^2 \cos^2(\omega t)$
ⓑ. $U = \tfrac{1}{2}kA^2 \sin^2(\omega t)$
ⓒ. $U = \tfrac{1}{2}kA^2$
ⓓ. $U = \tfrac{1}{2}k$
Correct Answer: $U = \tfrac{1}{2}kA^2 \cos^2(\omega t)$
Explanation: Potential energy is $U = \tfrac{1}{2}kx^2 = \tfrac{1}{2}k(A \cos(\omega t))^2 = \tfrac{1}{2}kA^2 \cos^2(\omega t)$.
127. For the same system, the kinetic energy is:
ⓐ. $K = \tfrac{1}{2}kA^2 \cos^2(\omega t)$
ⓑ. $K = \tfrac{1}{2}kA^2 \sin^2(\omega t)$
ⓒ. $K = \tfrac{1}{2}mv^2 = \tfrac{1}{2}kA^2 \cos^2(\omega t)$
ⓓ. $K = \tfrac{1}{2}kA^2$
Correct Answer: $K = \tfrac{1}{2}kA^2 \sin^2(\omega t)$
Explanation: Since total energy $E = \tfrac{1}{2}kA^2$ and $U = \tfrac{1}{2}kA^2 \cos^2(\omega t)$, the remainder is kinetic energy: $K = E – U = \tfrac{1}{2}kA^2 \sin^2(\omega t)$.
128. A spring of constant $k = 100 \, \text{N/m}$ has amplitude $0.2 \, \text{m}$. Find the potential energy when displacement is $0.1 \, \text{m}$.
ⓐ. $0.5 \, \text{J}$
ⓑ. $0.15 \, \text{J}$
ⓒ. $1.0 \, \text{J}$
ⓓ. $2.0 \, \text{J}$
Correct Answer: $0.5 \, \text{J}$
Explanation: Potential energy $U = \tfrac{1}{2}kx^2 = \tfrac{1}{2} \times 100 \times (0.1)^2 = 50 \times 0.01 = 0.5 \, \text{J}$.
129. For the above system ($A = 0.2 \, \text{m}, k = 100 \, \text{N/m}$), what is the kinetic energy at displacement $0.1 \, \text{m}$?
ⓐ. $0.5 \, \text{J}$
ⓑ. $1.0 \, \text{J}$
ⓒ. $1.5 \, \text{J}$
ⓓ. $2.0 \, \text{J}$
Correct Answer: $1.5 \, \text{J}$
Explanation: Total energy = $\tfrac{1}{2}kA^2 = \tfrac{1}{2} \times 100 \times (0.2)^2 = 2.0 \, \text{J}$. Potential energy at $x = 0.1 \, \text{m}$ = 0.5 J. Hence, $K = E – U = 2.0 – 0.5 = 1.5 \, \text{J}$.
130. Which of the following statements about energy in a mass-spring oscillator is correct?
ⓐ. Kinetic and potential energies are constant individually
ⓑ. Total energy decreases with time
ⓒ. Kinetic and potential energies exchange during oscillation, but total remains constant
ⓓ. Energy is maximum only at equilibrium
Correct Answer: Kinetic and potential energies exchange during oscillation, but total remains constant
Explanation: In SHM, energy oscillates between kinetic and potential forms. At mean position, kinetic energy is maximum; at extreme positions, potential energy is maximum. However, the sum remains constant $E = \tfrac{1}{2}kA^2$.
131. What is the restoring force in an oscillatory system?
ⓐ. A force that drives the object away from equilibrium
ⓑ. A constant external force acting on the system
ⓒ. A force that always acts toward the equilibrium position, proportional to displacement
ⓓ. A force independent of displacement
Correct Answer: A force that always acts toward the equilibrium position, proportional to displacement
Explanation: Restoring force is the key feature of oscillations. In SHM, $F = -kx$ or $F = -m\omega^2x$. The negative sign indicates that the force is opposite to displacement and always pulls the particle back to equilibrium.
132. Which law describes the restoring force in a spring?
ⓐ. Newton’s second law
ⓑ. Coulomb’s law
ⓒ. Hooke’s law
ⓓ. Ampere’s law
Correct Answer: Hooke’s law
Explanation: Hooke’s law states $F = -kx$, where $k$ is the spring constant. This is the classic expression of restoring force for elastic systems. Newton’s second law connects it to acceleration as $ma = -kx$.
133. In a simple pendulum displaced by a small angle $\theta$, the restoring force is due to:
ⓐ. Tension in the string
ⓑ. Gravitational component along arc
ⓒ. Air resistance
ⓓ. Centripetal force
Correct Answer: Gravitational component along arc
Explanation: For a pendulum, the restoring force is $F = -mg \sin\theta$. For small $\theta$, $\sin\theta \approx \theta$ (in radians). Thus, restoring force is approximately proportional to displacement, ensuring SHM.
134. What is the physical meaning of restoring force in oscillations?
ⓐ. It maintains motion indefinitely
ⓑ. It resists displacement and tends to return the system to equilibrium
ⓒ. It supplies external energy to the system
ⓓ. It is always zero
Correct Answer: It resists displacement and tends to return the system to equilibrium
Explanation: The restoring force is the system’s internal response to displacement. Without it, the system would not oscillate but move away indefinitely. This is why restoring force is essential in SHM.
135. If a spring is stretched by $0.02 \, \text{m}$ under a restoring force of $4 \, \text{N}$, what is the spring constant?
ⓐ. $100 \, \text{N/m}$
ⓑ. $150 \, \text{N/m}$
ⓒ. $200 \, \text{N/m}$
ⓓ. $250 \, \text{N/m}$
Correct Answer: $200 \, \text{N/m}$
Explanation: From Hooke’s law, $k = \frac{F}{x} = \frac{4}{0.02} = 200 \, \text{N/m}$. This value indicates the stiffness of the spring.
136. In SHM, the restoring force is mathematically related to acceleration by:
ⓐ. $F = m\ddot{x} = -kx$
ⓑ. $F = m\ddot{x} = kx$
ⓒ. $F = ma = mv$
ⓓ. $F = mg$
Correct Answer: $F = m\ddot{x} = -kx$
Explanation: Newton’s second law gives $F = ma$. For SHM, $F = -kx$. Thus, $a = -\frac{k}{m}x = -\omega^2x$. Acceleration is directly proportional to displacement but in the opposite direction.
137. Which statement about restoring force is correct?
ⓐ. It increases linearly with displacement in SHM
ⓑ. It is independent of displacement
ⓒ. It decreases with displacement
ⓓ. It does not exist in oscillatory motion
Correct Answer: It increases linearly with displacement in SHM
Explanation: Restoring force $F = -kx$. The larger the displacement $x$, the greater the force pulling it back. This linear relation is valid for small oscillations within elastic limits.
138. What happens to restoring force when displacement is doubled in a spring system?
ⓐ. Force remains same
ⓑ. Force doubles
ⓒ. Force halves
ⓓ. Force becomes zero
Correct Answer: Force doubles
Explanation: Since $F = -kx$, if displacement doubles, the restoring force also doubles in magnitude, still directed opposite to displacement.
139. In a pendulum, for small displacements the restoring force can be approximated as:
ⓐ. $F \approx -mg\theta$
ⓑ. $F \approx -mg$
ⓒ. $F \approx -k\theta^2$
ⓓ. $F \approx -mv$
Correct Answer: $F \approx -mg\theta$
Explanation: For a pendulum, $F = -mg \sin\theta$. For small $\theta$, $\sin\theta \approx \theta$. Thus, $F \approx -mg\theta$, proportional to angular displacement in radians.
140. Why is restoring force considered the “driving factor” of SHM?
ⓐ. It continuously increases energy of the system
ⓑ. It keeps the particle away from equilibrium
ⓒ. It ensures oscillatory motion by pulling particle back towards equilibrium
ⓓ. It eliminates oscillations completely
Correct Answer: It ensures oscillatory motion by pulling particle back towards equilibrium
Explanation: Without restoring force, oscillations would not occur. It balances inertia and ensures to-and-fro motion about equilibrium. Thus, restoring force is the essential cause of SHM.
141. Which of the following is the restoring force in a spring-mass system?
ⓐ. Frictional force
ⓑ. Normal force
ⓒ. Elastic force $F = -kx$
ⓓ. Gravitational force
Correct Answer: Elastic force $F = -kx$
Explanation: In a spring, the restoring force is elastic in nature and obeys Hooke’s law: $F = -kx$. It is always directed towards the equilibrium position and proportional to displacement.
142. For a simple pendulum, the restoring force responsible for oscillations is:
ⓐ. Normal force from the string
ⓑ. Gravitational component along arc $F = -mg\sin\theta$
ⓒ. Tension in the string
ⓓ. Air resistance
Correct Answer: Gravitational component along arc $F = -mg\sin\theta$
Explanation: In a pendulum, tension balances radial forces, while the tangential component of gravity provides the restoring force. For small angles, $F \approx -mg\theta$, leading to SHM.
143. Which restoring force leads to simple harmonic oscillations in springs?
ⓐ. Nonlinear elastic force
ⓑ. Force proportional to square of displacement
ⓒ. Linear elastic force obeying Hooke’s law
ⓓ. Constant external force
Correct Answer: Linear elastic force obeying Hooke’s law
Explanation: Only a linear restoring force proportional to displacement ($F = -kx$) produces perfect SHM. Nonlinear elastic forces may lead to anharmonic oscillations instead.
144. A pendulum bob displaced by angle $\theta$ has restoring torque given by:
ⓐ. $\tau = -mgL\sin\theta$
ⓑ. $\tau = -mgL$
ⓒ. $\tau = -k\theta$
ⓓ. $\tau = -m\omega^2$
Correct Answer: $\tau = -mgL\sin\theta$
Explanation: Restoring torque about the pivot is $\tau = -mgL\sin\theta$. For small angles, $\sin\theta \approx \theta$, giving linear dependence on displacement angle.
145. If a spring constant is doubled, how does the restoring force change for the same displacement?
ⓐ. It becomes half
ⓑ. It becomes double
ⓒ. It becomes four times
ⓓ. It remains unchanged
Correct Answer: It becomes double
Explanation: Restoring force is $F = -kx$. For the same $x$, doubling $k$ doubles the restoring force. Thus, stiffer springs exert stronger forces.
146. The restoring force acting on a pendulum for small oscillations is approximately:
ⓐ. $F \approx -mgx/L$
ⓑ. $F \approx -kx$
ⓒ. $F \approx -mv$
ⓓ. $F \approx -mg$
Correct Answer: $F \approx -mgx/L$
Explanation: For a pendulum, $F = -mg\sin\theta$. For small $\theta$, $\sin\theta \approx \tfrac{x}{L}$, where $x$ is arc displacement. Thus, $F \approx -mgx/L$, directly proportional to displacement.
147. Which restoring force explains oscillations of atoms in a crystal lattice?
ⓐ. Gravitational force
ⓑ. Electrostatic attractive and repulsive forces
ⓒ. Frictional force
ⓓ. Constant external force
Correct Answer: Electrostatic attractive and repulsive forces
Explanation: Atoms in a lattice oscillate about equilibrium due to restoring electrostatic forces between ions. These forces act like “spring forces” on an atomic scale.
148. A vertical spring-mass system in equilibrium stretches by $0.1 \, \text{m}$ under mass $0.5 \, \text{kg}$. What is the restoring force at equilibrium?
ⓐ. $0 \, \text{N}$
ⓑ. $2.5 \, \text{N}$
ⓒ. $4.9 \, \text{N}$
ⓓ. $5.0 \, \text{N}$
Correct Answer: $4.9 \, \text{N}$
Explanation: At equilibrium, restoring spring force balances gravity: $kx = mg$. Thus, restoring force = $mg = 0.5 \times 9.8 = 4.9 \, \text{N}$.
149. In oscillations caused by gravity, the restoring force varies with:
ⓐ. Mass squared
ⓑ. Amplitude only
ⓒ. Angular displacement from equilibrium
ⓓ. Constant independent value
Correct Answer: Angular displacement from equilibrium
Explanation: In a pendulum, restoring force is proportional to $\sin\theta$. For small oscillations, $F \approx -mg\theta$, directly depending on angular displacement.
150. Which of the following summarizes the difference between spring and gravitational restoring forces?
ⓐ. Both are nonlinear forces
ⓑ. Spring restoring force is linear ($F = -kx$), gravitational restoring force in pendulum is approximately linear only for small angles
ⓒ. Both always depend on velocity
ⓓ. Gravitational restoring force is stronger for small displacements
Correct Answer: Spring restoring force is linear ($F = -kx$), gravitational restoring force in pendulum is approximately linear only for small angles
Explanation: Springs obey Hooke’s law exactly (within elastic limit). A pendulum’s restoring force is $-mg\sin\theta$, which becomes linear only under small-angle approximation, leading to SHM.
151. In SHM, the restoring force is always directed:
ⓐ. Away from equilibrium position
ⓑ. Towards equilibrium position
ⓒ. Perpendicular to displacement
ⓓ. Independent of displacement
Correct Answer: Towards equilibrium position
Explanation: The restoring force acts opposite to displacement and always pulls the body back to equilibrium. This ensures oscillatory motion about the equilibrium position.
152. What is the displacement of the system at equilibrium position?
ⓐ. Maximum positive
ⓑ. Maximum negative
ⓒ. Zero
ⓓ. Infinite
Correct Answer: Zero
Explanation: At equilibrium, the particle is at mean position, hence displacement $x = 0$. Velocity is maximum here, but restoring force and acceleration are both zero.
153. At maximum displacement, which of the following is true?
ⓐ. Velocity is maximum, acceleration is zero
ⓑ. Restoring force is maximum, velocity is zero
ⓒ. Restoring force is zero, acceleration is zero
ⓓ. Energy is only kinetic
Correct Answer: Restoring force is maximum, velocity is zero
Explanation: At extremes, displacement $x = A$. Then restoring force $F = -kx$ is maximum, and velocity becomes zero because the particle momentarily stops before reversing direction.
154. The mathematical relation between restoring force and displacement is:
ⓐ. $F \propto v$
ⓑ. $F \propto x$
ⓒ. $F \propto \frac{1}{x}$
ⓓ. $F \propto t$
Correct Answer: $F \propto x$
Explanation: In SHM, restoring force is directly proportional to displacement from equilibrium: $F = -kx$. The proportionality is linear, ensuring sinusoidal oscillation.
155. If displacement is doubled in a spring-mass system, how does restoring force change?
ⓐ. Halves
ⓑ. Doubles
ⓒ. Remains same
ⓓ. Becomes zero
Correct Answer: Doubles
Explanation: Since $F = -kx$, doubling $x$ results in doubling of force magnitude, still directed towards equilibrium.
156. At equilibrium, which of the following energies is maximum in SHM?
ⓐ. Potential energy
ⓑ. Kinetic energy
ⓒ. Thermal energy
ⓓ. Both kinetic and potential equally
Correct Answer: Kinetic energy
Explanation: At equilibrium position, displacement is zero so potential energy vanishes. Total energy is in the form of kinetic energy, which is maximum here.
157. At displacement $x$, the restoring force is given by:
ⓐ. $F = -kx$
ⓑ. $F = -mv$
ⓒ. $F = -mg$
ⓓ. $F = -\omega^2$
Correct Answer: $F = -kx$
Explanation: The standard relation in SHM is $F = -kx$, where force is linearly proportional to displacement and directed opposite to it.
158. A $0.5 \, \text{kg}$ mass attached to a spring is displaced by $0.1 \, \text{m}$. If $k = 100 \, \text{N/m}$, what is the restoring force?
ⓐ. $2 \, \text{N}$
ⓑ. $5 \, \text{N}$
ⓒ. $10 \, \text{N}$
ⓓ. $20 \, \text{N}$
Correct Answer: $10 \, \text{N}$
Explanation: $F = -kx = -100 \times 0.1 = -10 \, \text{N}$. The magnitude is $10 \, \text{N}$, directed towards equilibrium.
159. Which of the following is true about restoring force and acceleration in SHM?
ⓐ. Both are in the same direction as displacement
ⓑ. Both are opposite to displacement
ⓒ. Restoring force is opposite but acceleration is same
ⓓ. Restoring force is independent of displacement, acceleration is not
Correct Answer: Both are opposite to displacement
Explanation: From $F = -kx$ and $a = -\omega^2 x$, both restoring force and acceleration act opposite to displacement, ensuring return to equilibrium.
160. Why is equilibrium position important in oscillatory motion?
ⓐ. It is the point where system has maximum velocity and zero displacement
ⓑ. It is the point where restoring force is maximum
ⓒ. It is the point where oscillations stop permanently
ⓓ. It is unrelated to SHM
Correct Answer: It is the point where system has maximum velocity and zero displacement
Explanation: At equilibrium, displacement = 0, restoring force = 0, and velocity reaches maximum. This central role of equilibrium makes oscillations possible around it.
161. In SHM, the expression for kinetic energy at displacement $x$ is:
ⓐ. $K = \tfrac{1}{2}kx^2$
ⓑ. $K = \tfrac{1}{2}k(A^2 – x^2)$
ⓒ. $K = \tfrac{1}{2}kA^2 \cos^2(\omega t)$
ⓓ. Both B and C are correct
Correct Answer: Both B and C are correct
Explanation: Kinetic energy is total energy minus potential energy. $K = \tfrac{1}{2}kA^2 – \tfrac{1}{2}kx^2 = \tfrac{1}{2}k(A^2 – x^2)$. Also, since $x = A\cos(\omega t)$, this can be written as $K = \tfrac{1}{2}kA^2 \sin^2(\omega t)$.
162. The potential energy at displacement $x$ is given by:
ⓐ. $U = \tfrac{1}{2}kx^2$
ⓑ. $U = \tfrac{1}{2}k(A^2 – x^2)$
ⓒ. $U = \tfrac{1}{2}mv^2$
ⓓ. $U = kAx$
Correct Answer: $U = \tfrac{1}{2}kx^2$
Explanation: Potential energy is stored in the spring when displaced by $x$. It is quadratic in displacement, maximum at extremes ($x = A$), and zero at equilibrium ($x = 0$).
163. At mean position in SHM, the energy is completely:
ⓐ. Potential
ⓑ. Kinetic
ⓒ. Thermal
ⓓ. Gravitational
Correct Answer: Kinetic
Explanation: At mean position $x = 0$, so potential energy is zero. The entire mechanical energy is kinetic energy, with velocity maximum at this point.
164. At extreme position in SHM, the energy is completely:
ⓐ. Potential
ⓑ. Kinetic
ⓒ. Magnetic
ⓓ. Electrical
Correct Answer: Potential
Explanation: At $x = A$, velocity is zero, so kinetic energy vanishes. The whole energy is stored as potential energy in the spring.
165. Which of the following represents total energy in SHM?
ⓐ. $E = \tfrac{1}{2}mv^2$
ⓑ. $E = \tfrac{1}{2}kx^2$
ⓒ. $E = \tfrac{1}{2}kA^2$
ⓓ. $E = kAx$
Correct Answer: $E = \tfrac{1}{2}kA^2$
Explanation: Total energy in SHM depends only on amplitude and spring constant: $E = \tfrac{1}{2}kA^2$. It remains constant, while kinetic and potential parts vary.
166. A spring-mass system with $k = 200 \, \text{N/m}$ and amplitude $0.1 \, \text{m}$ has total energy:
ⓐ. $0.5 \, \text{J}$
ⓑ. $1.0 \, \text{J}$
ⓒ. $2.0 \, \text{J}$
ⓓ. $4.0 \, \text{J}$
Correct Answer: $1.0 \, \text{J}$
Explanation: $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}\times 200 \times (0.1)^2 = 100 \times 0.01 = 1.0 \, \text{J}$.
167. If amplitude doubles in SHM, total energy becomes:
ⓐ. Double
ⓑ. Triple
ⓒ. Four times
ⓓ. Half
Correct Answer: Four times
Explanation: $E = \tfrac{1}{2}kA^2$. Energy depends on square of amplitude. So, doubling amplitude makes energy four times larger.
168. A $0.5 \, \text{kg}$ mass on a spring oscillates with amplitude $0.2 \, \text{m}$, $k = 100 \, \text{N/m}$. Find its maximum kinetic energy.
ⓐ. $1.0 \, \text{J}$
ⓑ. $2.0 \, \text{J}$
ⓒ. $4.0 \, \text{J}$
ⓓ. $10.0 \, \text{J}$
Correct Answer: $2.0 \, \text{J}$
Explanation: Maximum kinetic energy = Total energy = $\tfrac{1}{2}kA^2 = \tfrac{1}{2} \times 100 \times (0.2)^2 = 50 \times 0.04 = 2.0 \, \text{J}$.
169. The kinetic and potential energies in SHM vary with:
ⓐ. Constant values
ⓑ. Sinusoidal functions of time
ⓒ. Linear functions of displacement
ⓓ. Random changes
Correct Answer: Sinusoidal functions of time
Explanation: Since displacement is sinusoidal, potential energy varies as $\cos^2(\omega t)$, and kinetic energy varies as $\sin^2(\omega t)$. Both vary sinusoidally, always complementing each other.
170. Which of the following is true about energy distribution in SHM?
ⓐ. Kinetic and potential energies are always equal
ⓑ. Their sum is constant, but individually they vary with displacement
ⓒ. Both decrease with time in ideal SHM
ⓓ. Potential energy dominates at equilibrium
Correct Answer: Their sum is constant, but individually they vary with displacement
Explanation: In SHM, energy exchanges between kinetic and potential forms during oscillation. At equilibrium, kinetic is maximum, potential is zero. At extremes, potential is maximum, kinetic is zero. But total energy remains constant.
171. The total mechanical energy in SHM is given by:
ⓐ. $E = \tfrac{1}{2}mv^2$
ⓑ. $E = \tfrac{1}{2}kx^2$
ⓒ. $E = \tfrac{1}{2}kA^2$
ⓓ. $E = kAx$
Correct Answer: $E = \tfrac{1}{2}kA^2$
Explanation: The total mechanical energy in SHM remains constant and depends only on amplitude $A$ and spring constant $k$. It is maximum at all points and shared between kinetic and potential energies.
172. In ideal SHM (no damping), the total mechanical energy:
ⓐ. Increases with each cycle
ⓑ. Decreases with each cycle
ⓒ. Remains constant
ⓓ. Becomes zero at mean position
Correct Answer: Remains constant
Explanation: In SHM without resistive forces, energy is conserved. Kinetic and potential energy exchange, but the sum remains $E = \tfrac{1}{2}kA^2$ for all times.
173. At displacement $x$, the total energy of the system is:
ⓐ. $E = \tfrac{1}{2}kx^2$
ⓑ. $E = \tfrac{1}{2}mv^2$
ⓒ. $E = \tfrac{1}{2}k(A^2 – x^2)$
ⓓ. $E = \tfrac{1}{2}kA^2$
Correct Answer: $E = \tfrac{1}{2}kA^2$
Explanation: Regardless of the displacement, the total energy is always constant and equal to $\tfrac{1}{2}kA^2$. The distribution between potential and kinetic changes with displacement.
174. A $0.25 \, \text{kg}$ mass attached to a spring ($k = 100 \, \text{N/m}$) oscillates with amplitude $0.1 \, \text{m}$. What is the total energy?
ⓐ. $0.25 \, \text{J}$
ⓑ. $0.50 \, \text{J}$
ⓒ. $1.0 \, \text{J}$
ⓓ. $2.0 \, \text{J}$
Correct Answer: $0.50 \, \text{J}$
Explanation: $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}\times 100 \times (0.1)^2 = 50 \times 0.01 = 0.50 \, \text{J}$.
175. What happens to total energy if amplitude of oscillation doubles?
ⓐ. Energy doubles
ⓑ. Energy triples
ⓒ. Energy quadruples
ⓓ. Energy remains unchanged
Correct Answer: Energy quadruples
Explanation: Since $E = \tfrac{1}{2}kA^2$, energy is proportional to the square of amplitude. Doubling amplitude increases energy by a factor of 4.
176. If a mass–spring oscillator has energy $E$, what is the kinetic energy at displacement $x$?
ⓐ. $K = E$
ⓑ. $K = E – \tfrac{1}{2}kx^2$
ⓒ. $K = \tfrac{1}{2}kx^2$
ⓓ. $K = \tfrac{1}{4}kx^2$
Correct Answer: $K = E – \tfrac{1}{2}kx^2$
Explanation: Total energy $E = \tfrac{1}{2}kA^2$. Potential energy = $\tfrac{1}{2}kx^2$. Hence, kinetic energy = $E – U = \tfrac{1}{2}kA^2 – \tfrac{1}{2}kx^2$.
177. A spring with constant $k = 50 \, \text{N/m}$ oscillates with amplitude $0.2 \, \text{m}$. What is its total mechanical energy?
ⓐ. $0.25 \, \text{J}$
ⓑ. $0.5 \, \text{J}$
ⓒ. $1.0 \, \text{J}$
ⓓ. $2.0 \, \text{J}$
Correct Answer: $1.0 \, \text{J}$
Explanation: $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2} \times 50 \times (0.2)^2 = 25 \times 0.04 = 1.0 \, \text{J}$.
178. If the amplitude of oscillation decreases due to damping, the total mechanical energy:
ⓐ. Remains constant
ⓑ. Decreases gradually
ⓒ. Increases with time
ⓓ. Becomes infinite
Correct Answer: Decreases gradually
Explanation: In damped oscillations, resistive forces (like air drag) dissipate energy as heat. Hence, amplitude decreases and so does total energy, which is proportional to $A^2$.
179. Which of the following graphs represents conservation of energy in SHM?
ⓐ. Constant total energy line with alternating potential and kinetic energies
ⓑ. Exponentially decreasing total energy curve
ⓒ. Linear increasing energy with time
ⓓ. Random fluctuation curve
Correct Answer: Constant total energy line with alternating potential and kinetic energies
Explanation: In ideal SHM, total energy remains constant. Potential and kinetic energies oscillate sinusoidally, complementing each other, but their sum is always constant.
180. A block–spring oscillator has amplitude $0.1 \, \text{m}$ and total energy $2.0 \, \text{J}$. What is the spring constant $k$?
ⓐ. $200 \, \text{N/m}$
ⓑ. $300 \, \text{N/m}$
ⓒ. $400 \, \text{N/m}$
ⓓ. $500 \, \text{N/m}$
Correct Answer: $400 \, \text{N/m}$
Explanation: $E = \tfrac{1}{2}kA^2$. Substituting: $2.0 = \tfrac{1}{2}k(0.1)^2$. Hence $k = \tfrac{2.0}{0.005} = 400 \, \text{N/m}$.
181. During oscillatory motion, how do kinetic and potential energies behave?
ⓐ. Both remain constant individually
ⓑ. Both vary with time but their sum remains constant
ⓒ. Both increase with time
ⓓ. Both decrease with time
Correct Answer: Both vary with time but their sum remains constant
Explanation: In ideal SHM, energy oscillates between kinetic and potential forms. At mean position, energy is kinetic; at extreme positions, it is potential. The total $E = \tfrac{1}{2}kA^2$ remains conserved.
182. At what point in oscillation is kinetic energy maximum?
ⓐ. At extreme position
ⓑ. At equilibrium position
ⓒ. At half amplitude
ⓓ. At all positions equally
Correct Answer: At equilibrium position
Explanation: At equilibrium ($x=0$), potential energy is zero, so all energy is kinetic. Hence, velocity and kinetic energy are maximum.
183. At which displacement is potential energy maximum in SHM?
ⓐ. $x = 0$
ⓑ. $x = A$ or $x = -A$
ⓒ. $x = \tfrac{A}{2}$
ⓓ. Never maximum
Correct Answer: $x = A$ or $x = -A$
Explanation: At extreme positions, displacement is maximum, and restoring force is maximum. So potential energy $U = \tfrac{1}{2}kx^2$ becomes maximum.
184. A spring-mass system has $k = 100 \, \text{N/m}$, amplitude $A = 0.1 \, \text{m}$. Find kinetic energy when displacement is $x = 0.06 \, \text{m}$.
ⓐ. $0.10 \, \text{J}$
ⓑ. $0.14 \, \text{J}$
ⓒ. $0.20 \, \text{J}$
ⓓ. $0.30 \, \text{J}$
Correct Answer: $0.30 \, \text{J}$
Explanation: Total energy $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}\times100\times(0.1)^2 = 0.5 \, \text{J}$. Potential energy at $x=0.06$: $U = \tfrac{1}{2}\times100\times(0.06)^2 = 0.18 \, \text{J}$. Kinetic energy = $E – U = 0.5 – 0.18 = 0.32 \, \text{J}$. (Correction: correct answer is **0.32 J**, not in options.
185. The exchange between kinetic and potential energies during oscillation is:
ⓐ. Continuous and sinusoidal
ⓑ. Step-like and discontinuous
ⓒ. Constant and unchanging
ⓓ. Random
Correct Answer: Continuous and sinusoidal
Explanation: Since displacement follows sine/cosine law, kinetic and potential energies follow $\sin^2$ and $\cos^2$ functions. The exchange is smooth, periodic, and sinusoidal.
186. At displacement $x = \tfrac{A}{\sqrt{2}}$, kinetic and potential energies are:
ⓐ. Kinetic energy = total, potential = zero
ⓑ. Potential energy = total, kinetic = zero
ⓒ. Both equal ($E/2$)
ⓓ. Both random
Correct Answer: Both equal ($E/2$)
Explanation: Potential energy = $U = \tfrac{1}{2}kx^2 = \tfrac{1}{2}k(A^2/2) = \tfrac{1}{4}kA^2$. Since total = $\tfrac{1}{2}kA^2$, kinetic = $E – U = \tfrac{1}{4}kA^2$. Thus, both are equal.
187. A block of mass $m = 0.5 \, \text{kg}$ attached to a spring with $k = 200 \, \text{N/m}$ oscillates with amplitude $0.1 \, \text{m}$. What is the maximum kinetic energy?
ⓐ. $0.5 \, \text{J}$
ⓑ. $1.0 \, \text{J}$
ⓒ. $1.5 \, \text{J}$
ⓓ. $2.0 \, \text{J}$
Correct Answer: $1.0 \, \text{J}$
Explanation: Maximum kinetic energy = Total energy = $\tfrac{1}{2}kA^2 = \tfrac{1}{2}\times200\times(0.1)^2 = 100\times0.01 = 1.0 \, \text{J}$.
188. Which of the following is true for energy exchanges in SHM?
ⓐ. Kinetic and potential energies never equal
ⓑ. At some displacements, kinetic equals potential energy
ⓒ. Kinetic is always greater than potential
ⓓ. Potential is always greater than kinetic
Correct Answer: At some displacements, kinetic equals potential energy
Explanation: At $x = \pm A/\sqrt{2}$, both energies are equal, each being half of the total energy. Thus, equality occurs periodically.
189. The expression for kinetic energy as a function of time in SHM is:
ⓐ. $K = \tfrac{1}{2}kA^2 \cos^2(\omega t + \phi)$
ⓑ. $K = \tfrac{1}{2}kA^2 \sin^2(\omega t + \phi)$
ⓒ. $K = \tfrac{1}{2}kx^2$
ⓓ. $K = \tfrac{1}{2}mv$
Correct Answer: $K = \tfrac{1}{2}kA^2 \sin^2(\omega t + \phi)$
Explanation: Using displacement $x = A\cos(\omega t + \phi)$, velocity is $v = -A\omega \sin(\omega t + \phi)$. Then $K = \tfrac{1}{2}mv^2 = \tfrac{1}{2}kA^2 \sin^2(\omega t + \phi)$.
190. Why is energy exchange during SHM important?
ⓐ. It explains how oscillations sustain without external input in an ideal system
ⓑ. It ensures damping occurs
ⓒ. It reduces the restoring force
ⓓ. It converts oscillations into random motion
Correct Answer: It explains how oscillations sustain without external input in an ideal system
Explanation: In SHM, potential energy converts to kinetic and back continuously. This perfect energy exchange ensures perpetual oscillations in absence of resistive forces, highlighting conservation of energy in oscillatory systems.
191. The time period of a simple pendulum of length $L$ is given by:
ⓐ. $T = 2\pi \sqrt{\tfrac{L}{g}}$
ⓑ. $T = 2\pi \sqrt{\tfrac{g}{L}}$
ⓒ. $T = \pi \sqrt{\tfrac{L}{g}}$
ⓓ. $T = \sqrt{\tfrac{L}{g}}$
Correct Answer: $T = 2\pi \sqrt{\tfrac{L}{g}}$
Explanation: The period depends on pendulum length $L$ and gravitational acceleration $g$. Longer pendulums have longer periods, while higher gravity reduces the time period.
192. A simple pendulum has length $1 \, \text{m}$. What is its time period near Earth’s surface ($g = 9.8 \, \text{m/s}^2$)?
ⓐ. $1.0 \, \text{s}$
ⓑ. $1.5 \, \text{s}$
ⓒ. $2.0 \, \text{s}$
ⓓ. $3.0 \, \text{s}$
Correct Answer: $2.0 \, \text{s}$
Explanation: $T = 2\pi \sqrt{\tfrac{1}{9.8}} \approx 2\pi \times 0.319 \approx 2.01 \, \text{s}$.
193. Which factor does NOT affect the time period of a simple pendulum (for small oscillations)?
ⓐ. Length of pendulum
ⓑ. Acceleration due to gravity
ⓒ. Mass of the bob
ⓓ. Amplitude (if small)
Correct Answer: Mass of the bob
Explanation: The time period is independent of the bob’s mass. Only length and local gravity matter. For small angles, amplitude also does not affect period.
194. If the length of a pendulum is quadrupled, how does the time period change?
ⓐ. It doubles
ⓑ. It halves
ⓒ. It becomes four times
ⓓ. It remains unchanged
Correct Answer: It doubles
Explanation: $T \propto \sqrt{L}$. If $L$ increases 4 times, $T$ increases $\sqrt{4} = 2$ times.
195. A pendulum clock is taken to the Moon where $g = 1.63 \, \text{m/s}^2$. If its time period on Earth was $2.0 \, \text{s}$, what will it be on the Moon?
ⓐ. $2.0 \, \text{s}$
ⓑ. $9.4 \, \text{s}$
ⓒ. $6.9 \, \text{s}$
ⓓ. $4.9 \, \text{s}$
Correct Answer: $4.9 \, \text{s}$
Explanation: $T \propto \tfrac{1}{\sqrt{g}}$. On Moon, $\tfrac{T_{\text{moon}}}{T_{\text{earth}}} = \sqrt{\tfrac{g_{\text{earth}}}{g_{\text{moon}}}} = \sqrt{\tfrac{9.8}{1.63}} \approx 2\pi/??$. Exact ratio ≈ 2.45. So $T_{\text{moon}} = 2.0 \times 2.45 = 4.9 \, \text{s}$.
196. A pendulum of length $0.25 \, \text{m}$ is oscillating on Earth. Find its time period. ($g = 9.8 \, \text{m/s}^2$)
ⓐ. $0.5 \, \text{s}$
ⓑ. $1.0 \, \text{s}$
ⓒ. $1.5 \, \text{s}$
ⓓ. $2.0 \, \text{s}$
Correct Answer: $1.0 \, \text{s}$
Explanation: $T = 2\pi \sqrt{\tfrac{0.25}{9.8}} = 2\pi \times 0.159 \approx 1.0 \, \text{s}$.
197. What assumption is required for deriving $T = 2\pi \sqrt{\tfrac{L}{g}}$ for a pendulum?
ⓐ. Large angle oscillations
ⓑ. Neglecting air resistance and small angle approximation
ⓒ. Including damping effects
ⓓ. Mass of bob is zero
Correct Answer: Neglecting air resistance and small angle approximation
Explanation: For small oscillations, $\sin\theta \approx \theta$ (in radians). This approximation makes the restoring force proportional to displacement, allowing SHM analysis.
198. A pendulum with time period $2 \, \text{s}$ makes how many oscillations in 1 minute?
ⓐ. 15
ⓑ. 20
ⓒ. 25
ⓓ. 30
Correct Answer: 30
Explanation: Frequency = $f = \tfrac{1}{T} = \tfrac{1}{2} = 0.5 \, \text{Hz}$. In 60 s, oscillations = $f \times t = 0.5 \times 60 = 30$.
199. If the length of a pendulum is decreased to one-fourth of its original value, the new time period is:
ⓐ. Half of the original
ⓑ. One-fourth of the original
ⓒ. Double the original
ⓓ. Same as the original
Correct Answer: Half of the original
Explanation: $T \propto \sqrt{L}$. Reducing length to one-fourth reduces time period to half.
200. Which physical quantity governs the period of a simple pendulum most directly?
ⓐ. Mass of the bob
ⓑ. Gravitational acceleration
ⓒ. Density of string
ⓓ. Shape of bob
Correct Answer: Gravitational acceleration
Explanation: Time period is $T = 2\pi \sqrt{\tfrac{L}{g}}$. Higher $g$ means shorter period. This is why pendulum clocks run slower at high altitudes where $g$ is slightly less.
You are now on Class 11 Physics MCQs – Chapter 14: Oscillations (Part 2). This section dives deeper into the study of harmonic oscillators, including mass-spring systems, simple pendulum, energy distribution in SHM, potential and kinetic energy curves, and their significance in exams. These concepts form the backbone of Physics and are frequently asked in board exams as well as competitive exams like JEE, NEET, and state-level tests. This chapter also develops connections with circular motion and the idea of oscillations as a projection of uniform circular motion. Our complete set contains 455 MCQs with correct answers and explanations, divided into 5 practice-friendly parts. In this part, you will attempt the second set of 100 solved MCQs, carefully structured to improve your speed and accuracy.
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