Simple and Physical Pendulums - College Physics

Two physical pendulums are made from identical uniform rods of length $L$ and mass $m$ . Pendulum X pivots about one end, while pendulum Y pivots about a point $L/4$ from one end. What is the ratio of their periods $T_X/T_Y$ ?
Answer: $\sqrt{\frac{8}{7}}$ For pendulum X (pivot at end): $I_X = \frac{1}{3}mL^2$ , $h_X = L/2$ , so $T_X = 2\pi\sqrt{\frac{\frac{1}{3}mL^2}{mg \cdot \frac{L}{2}}} = 2\pi\sqrt{\frac{2L}{3g}}$ . For pendulum Y (pivot at $L/4$ from end): Using parallel axis theorem, $I_Y = \frac{1}{12}mL^2 + m(\frac{L}{4})^2 = \frac{1}{12}mL^2 + \frac{1}{16}mL^2 = \frac{7}{48}mL^2$ , and $h_Y = |\frac{L}{2} - \frac{L}{4}| = \frac{L}{4}$ . So $T_Y = 2\pi\sqrt{\frac{\frac{7}{48}mL^2}{mg \cdot \frac{L}{4}}} = 2\pi\sqrt{\frac{7L}{12g}}$ . Therefore: $\frac{T_X}{T_Y} = \sqrt{\frac{\frac{2L}{3g}}{\frac{7L}{12g}}} = \sqrt{\frac{2 \cdot 12}{3 \cdot 7}} = \sqrt{\frac{8}{7}}$ .
A uniform rod of length $L$ and mass $m$ is pivoted at a point located at distance $d$ from one end, where $d < L/2$ . For small oscillations as a physical pendulum, which expression correctly represents the period?
Answer: $T = 2\pi\sqrt{\frac{mL^2/12 + m(L/2 - d)^2}{mg(L/2 - d)}}$ For a physical pendulum, $T = 2\pi\sqrt{\frac{I}{mgh}}$ where $I$ is the moment of inertia about the pivot, $h$ is the distance from pivot to center of mass. The center of mass of a uniform rod is at $L/2$ from either end. Since the pivot is at distance $d$ from one end, $h = |L/2 - d| = L/2 - d$ (since $d < L/2$ ). Using parallel axis theorem: $I = I_{cm} + mh^2 = \frac{mL^2}{12} + m(L/2 - d)^2$ . Choice A uses wrong distance to CM. Choice B has correct $h$ but wrong $I_{cm}$ term. Choice C uses wrong distance to CM in denominator.
A simple pendulum oscillates with period $T_0$ when the amplitude is small. If the supporting string is replaced with a rigid rod of the same length and negligible mass, and the bob is treated as a point mass, what happens to the period?
Answer: The period remains $T_0$ since the bob is still a point mass at the same distance For a simple pendulum with a point mass bob, the period depends only on the length $L$ and gravitational acceleration: $T = 2\pi\sqrt{L/g}$ . Whether the bob is connected by a string or a massless rigid rod doesn't affect this relationship, as long as the distance from pivot to bob remains $L$ and the bob can still be treated as a point mass. The key insight is that both configurations produce the same restoring torque $\tau = -mgL\sin\theta$ for small angles. Choice A incorrectly suggests mechanical constraints affect the period. Choice B confuses this with coupled oscillator systems. Choice D is wrong because rigid rods can certainly support pendulum motion (like a meter stick pendulum).
A simple pendulum with length L = 0.80 m oscillates with period T = 1.8 s in a location with unknown gravitational acceleration. A second pendulum with length 1.20 m is placed at the same location. What is the period of the second pendulum?
Answer: $T_2 = 1.8 \times \sqrt{\frac{1.20}{0.80}} = 2.2$ s When you encounter pendulum problems, remember that the period depends only on length and gravitational acceleration, following the formula $T = 2\pi\sqrt{\frac{L}{g}}$ . Since both pendulums are at the same location, they experience the same gravitational acceleration, allowing you to compare their periods directly. To find the relationship between the two periods, you can set up a ratio. For the first pendulum: $T_1 = 2\pi\sqrt{\frac{L_1}{g}}$ , and for the second: $T_2 = 2\pi\sqrt{\frac{L_2}{g}}$ . Taking the ratio $\frac{T_2}{T_1} = \sqrt{\frac{L_2}{L_1}}$ , you get $T_2 = T_1 \sqrt{\frac{L_2}{L_1}} = 1.8 \times \sqrt{\frac{1.20}{0.80}} = 2.2$ s. Choice A correctly applies this square root relationship. Choice B uses a linear relationship ( $\frac{L_2}{L_1}$ ), which would be correct for simple proportional relationships but ignores that period depends on the square root of length. Choice C flips the length ratio, which would give you the period if the second pendulum were shorter than the first. Choice D squares the length ratio ( $\left(\frac{L_2}{L_1}\right)^2$ ), representing a common algebraic error when students confuse which operations involve square roots. The key insight is recognizing that pendulum period scales with the square root of length, not linearly. When comparing pendulums at the same location, always use $\frac{T_2}{T_1} = \sqrt{\frac{L_2}{L_1}}$ to avoid the most common mistakes in pendulum problems.
A grandfather clock uses a simple pendulum with a period of exactly 2.000 s at 20°C. The pendulum rod is made of steel with coefficient of linear expansion α = 11 × 10⁻⁶ /°C. On a hot day when the temperature rises to 35°C, how much time will the clock gain or lose per day?
Answer: The clock will lose 7.1 seconds per day This problem tests thermal expansion effects on pendulum timing. When you encounter pendulum problems involving temperature changes, remember that the period depends on length, and thermal expansion changes that length. For a simple pendulum, the period is $T = 2\pi\sqrt{\frac{L}{g}}$ . When temperature increases, the steel rod expands according to $L_{new} = L_0(1 + \alpha\Delta T)$ , where $\alpha = 11 \times 10^{-6}/°C$ and $\Delta T = 35°C - 20°C = 15°C$ . The fractional change in length is $\frac{\Delta L}{L_0} = \alpha\Delta T = (11 \times 10^{-6})(15) = 1.65 \times 10^{-4}$ . Since $T \propto \sqrt{L}$ , the fractional change in period is $\frac{\Delta T}{T_0} = \frac{1}{2} \times 1.65 \times 10^{-4} = 8.25 \times 10^{-5}$ . The new period becomes $T_{new} = 2.000(1 + 8.25 \times 10^{-5}) = 2.000165$ s. The clock now runs slower, losing $0.000165$ s every swing. In 24 hours, there are $\frac{24 \times 3600}{2.000} = 43,200$ swings, so the total time lost is $(0.000165)(43,200) = 7.1$ seconds. Answer A is correct: the clock loses 7.1 seconds per day. Answer B doubles this value incorrectly. Answers C and D wrongly suggest the clock gains time—expansion makes the pendulum longer and slower, not faster. Study tip: For thermal expansion problems with pendulums, remember that longer means slower. The fractional period change is always half the fractional length change due to the square root relationship.
A physical pendulum consists of a thin ring of radius $R$ and mass $m$ suspended from a point on its circumference. What is the period of small oscillations, and how does it compare to a simple pendulum of length $R$ ?
Answer: $T = 2\pi\sqrt{\frac{2R}{g}}$ ; this is $\sqrt{2}$ times longer than the simple pendulum For a thin ring: $I_{cm} = mR^2$ . Using parallel axis theorem with pivot on circumference: $I = I_{cm} + mR^2 = mR^2 + mR^2 = 2mR^2$ . The distance from pivot to center of mass is $h = R$ . Therefore: $T = 2\pi\sqrt{\frac{I}{mgh}} = 2\pi\sqrt{\frac{2mR^2}{mgR}} = 2\pi\sqrt{\frac{2R}{g}}$ . A simple pendulum of length $R$ has period $T_{simple} = 2\pi\sqrt{\frac{R}{g}}$ . The ratio is $\frac{T}{T_{simple}} = \sqrt{2}$ , so the physical pendulum is $\sqrt{2}$ times longer. Choice B incorrectly ignores the distributed mass. Choice C uses wrong moment of inertia. Choice D has an incorrect factor of $2\pi$ .
A simple pendulum has length $L$ and oscillates with small amplitude in a vertical plane. If the pendulum is now placed in a uniform horizontal electric field $E$ , and the bob has charge $+q$ , what is the new period when the bob oscillates about its new equilibrium position?
Answer: $T = 2\pi\sqrt{\frac{L}{\sqrt{g^2 + (qE/m)^2}}}$ The electric field creates a horizontal force $F_E = qE$ on the charged bob. In equilibrium, the bob hangs at angle $\theta_0$ where $\tan\theta_0 = \frac{qE}{mg}$ . The effective gravity is $g_{eff} = \sqrt{g^2 + (qE/m)^2}$ pointing toward the equilibrium position. For small oscillations about this new equilibrium, the restoring force per unit displacement follows the same form as a simple pendulum but with $g_{eff}$ replacing $g$ . Choice A ignores the electric field. Choice C incorrectly includes the equilibrium angle but uses original $g$ . Choice D incorrectly adds the accelerations rather than finding the magnitude of the resultant.
A physical pendulum consists of a uniform disk of radius $R$ that can swing about a horizontal axis located at distance $R/2$ from the disk's center. What is the length of a simple pendulum that would have the same period as this physical pendulum?
Answer: $\frac{3R}{4}$ For the physical pendulum: $I = I_{cm} + mh^2 = \frac{1}{2}mR^2 + m(\frac{R}{2})^2 = \frac{1}{2}mR^2 + \frac{1}{4}mR^2 = \frac{3}{4}mR^2$ . The distance from pivot to center of mass is $h = R/2$ . So $T = 2\pi\sqrt{\frac{I}{mgh}} = 2\pi\sqrt{\frac{\frac{3}{4}mR^2}{mg \cdot \frac{R}{2}}} = 2\pi\sqrt{\frac{3R}{2g}}$ . For a simple pendulum with the same period: $T = 2\pi\sqrt{\frac{L_{equiv}}{g}}$ . Setting equal: $\frac{3R}{2g} = \frac{L_{equiv}}{g}$ , so $L_{equiv} = \frac{3R}{2}$ . Wait, this should be $\sqrt{\frac{3R}{2g}} = \sqrt{\frac{L_{equiv}}{g}}$ , giving $L_{equiv} = \frac{3R}{2} \cdot \frac{1}{2} = \frac{3R}{4}$ .
A simple pendulum of length $L$ oscillates with maximum angular displacement $\theta_0$ . At the moment when the bob passes through the bottom of its swing, the string suddenly breaks. Immediately after the string breaks, what is the magnitude of the bob's acceleration?
Answer: $g$ When analyzing pendulum motion that suddenly changes, you need to carefully distinguish between the forces acting before and after the disruption occurs. At the bottom of the pendulum's swing, the bob is moving in a circular path with some speed $v$ . During normal pendulum motion, two forces act on the bob: the gravitational force $mg$ downward and the tension force $T$ upward along the string. The net upward force provides the centripetal acceleration needed for circular motion: $T - mg = \frac{mv^2}{L}$ . The instant the string breaks, however, the tension force completely disappears. Now only gravity acts on the bob, so the acceleration immediately becomes $g$ downward. The bob's velocity doesn't change instantaneously (it still has the same horizontal velocity it had at the bottom), but its acceleration changes abruptly from the centripetal value to pure gravitational acceleration. Choice A ( $0$ ) incorrectly assumes the bob somehow stops accelerating. Choice B ( $g\cos\theta_0$ ) mistakenly applies the component of gravity along the string at maximum displacement, which isn't relevant once the string breaks. Choice C ( $g(3 - 2\cos\theta_0)$ ) appears to be the centripetal acceleration during normal pendulum motion, but this becomes irrelevant the moment the string disappears. Choice D ( $g$ ) correctly recognizes that only gravity acts on a free-falling object. Remember: when constraints suddenly disappear in mechanics problems, immediately reassess what forces remain acting on the object. Don't carry over force relationships from the constrained motion.
A simple pendulum oscillates in an elevator. When the elevator accelerates upward at 2.0 m/s², the period decreases from 2.0 s to 1.8 s. What was the original length of the pendulum when the elevator was at rest?
Answer: L = 1.0 m When you encounter pendulum problems in accelerating reference frames, remember that the effective gravitational acceleration changes, which directly affects the period through the pendulum equation. For a simple pendulum, the period is given by $T = 2\pi\sqrt{\frac{L}{g_{eff}}}$ , where $g_{eff}$ is the effective gravitational acceleration. When the elevator is at rest, $g_{eff} = g = 9.8 \text{ m/s}^2$ . When accelerating upward at $a = 2.0 \text{ m/s}^2$ , the effective gravity becomes $g_{eff} = g + a = 11.8 \text{ m/s}^2$ . Using the period equation for both situations: - At rest: $2.0 = 2\pi\sqrt{\frac{L}{9.8}}$ - Accelerating: $1.8 = 2\pi\sqrt{\frac{L}{11.8}}$ From the first equation: $L = \frac{(2.0)^2 \times 9.8}{4\pi^2} = \frac{39.2}{39.48} \approx 1.0 \text{ m}$ You can verify this with the second equation: $L = \frac{(1.8)^2 \times 11.8}{4\pi^2} \approx 1.0 \text{ m}$ . This confirms answer B. Choice A (0.81 m) would give periods that are too short for both scenarios. Choice C (1.2 m) would yield a rest period of about 2.2 s, not 2.0 s. Choice D (1.4 m) would produce an even longer rest period of approximately 2.4 s. Study tip: In accelerating reference frame problems, always adjust the effective gravity first ( $g_{eff} = g \pm a$ ), then apply your standard formulas. Upward acceleration increases effective gravity, making pendulums swing faster.
A uniform meter stick is suspended as a physical pendulum from a point $30 \text{ cm}$ from one end. If the period of small oscillations is measured to be $T$ , what would be the period if the same meter stick were suspended from a point $20 \text{ cm}$ from the center of the stick?
Answer: The period would be the same $T$ This demonstrates the key property of physical pendulums: there are two pivot points that give the same period. For the first case, the pivot is $30 \text{ cm}$ from one end, which is $20 \text{ cm}$ from the center (since the stick is $100 \text{ cm}$ long). For the second case, the pivot is $20 \text{ cm}$ from the center. Both configurations have the same distance $h = 20 \text{ cm}$ from pivot to center of mass. Since $T = 2\pi\sqrt{\frac{I}{mgh}} = 2\pi\sqrt{\frac{I_{cm} + mh^2}{mgh}}$ , and both have the same $h$ , they have identical periods. This is related to the concept of compound pendulum equivalence. Choice D is incorrect because the period formula is independent of mass $m$ .
A simple pendulum oscillates with an amplitude of 15°. If the amplitude is reduced to 5°, how does the period change?
Answer: The period decreases by approximately 1% due to reduced nonlinear effects When analyzing pendulum motion, you need to distinguish between the idealized simple harmonic motion formula and real-world nonlinear effects that depend on amplitude. The standard pendulum period formula $T = 2\pi\sqrt{\frac{L}{g}}$ assumes small angles where $\sin\theta \approx \theta$ . However, for larger amplitudes, nonlinear effects become significant. The actual period is longer than the idealized formula predicts, and this deviation increases with amplitude. For small but finite amplitudes, the period can be approximated as $T \approx 2\pi\sqrt{\frac{L}{g}}\left(1 + \frac{\theta_0^2}{16}\right)$ , where $\theta_0$ is the amplitude in radians. When amplitude decreases from 15° to 5°, the nonlinear correction term becomes much smaller, so the actual period decreases toward the ideal value by approximately 1%. Option A incorrectly states the period increases with reduced nonlinear effects. Since nonlinear effects make the period longer than ideal, reducing them decreases the period. Options C and D both incorrectly assume the period scales proportionally with amplitude. This fundamental misunderstanding ignores that the basic pendulum period formula is independent of amplitude for small oscillations. The period changes due to nonlinear corrections, not direct proportionality. The correct answer is B because reducing amplitude from 15° to 5° decreases the nonlinear effects that were making the period longer than the idealized value. Study tip: Remember that nonlinear effects in pendulums always increase the period beyond the ideal $T = 2\pi\sqrt{L/g}$ formula, so reducing amplitude brings you closer to this shorter ideal period.
A simple pendulum has a length of 1.0 m and oscillates with small amplitude. If the pendulum bob is replaced with one having twice the mass, what happens to the period of oscillation?
Answer: The period remains the same When you encounter pendulum problems, focus on what variables actually appear in the period formula. This tests your understanding of which physical quantities affect oscillatory motion. For a simple pendulum with small amplitude oscillations, the period is given by $T = 2\pi\sqrt{\frac{L}{g}}$ , where $L$ is the length and $g$ is gravitational acceleration. Notice that mass ( $m$ ) doesn't appear anywhere in this equation. This means the period depends only on the pendulum's length and the local gravitational field strength, not on the mass of the bob. When you double the mass from $m$ to $2m$ , the gravitational force on the bob doubles ( $F = mg$ becomes $F = 2mg$ ). However, the bob's inertia also doubles, since inertia is directly proportional to mass. These effects exactly cancel each other out: the increased gravitational restoring force is perfectly balanced by the increased resistance to acceleration. Therefore, the period remains unchanged. Looking at the wrong answers: (A) suggests the period increases by $\sqrt{2}$ , which would happen if you incorrectly thought mass appeared in the numerator under the square root. (B) implies a direct proportional relationship between mass and period, ignoring the inertial effects entirely. (D) suggests the period decreases by $\sqrt{2}$ , which might come from incorrectly placing mass in the denominator. Remember this key insight: for simple harmonic oscillators, when both the restoring force and inertia change by the same factor, the period stays constant. This principle applies to many oscillatory systems beyond just pendulums.
A simple pendulum is observed to complete exactly 50 oscillations in 100.0 seconds. If the length is increased by 21%, how long will it take to complete 50 oscillations?
Answer: 110.0 seconds This question tests your understanding of how pendulum period depends on length. When you encounter pendulum problems, remember that the period (time for one complete oscillation) follows the relationship $T = 2\pi\sqrt{\frac{L}{g}}$ , where L is length and g is gravitational acceleration. First, let's find the original period. With 50 oscillations in 100.0 seconds, each oscillation takes $T_1 = \frac{100.0}{50} = 2.0$ seconds. When length increases by 21%, the new length becomes $L_2 = 1.21L_1$ . Since period is proportional to the square root of length, the new period is: $T_2 = T_1\sqrt{\frac{L_2}{L_1}} = 2.0\sqrt{1.21} = 2.0 \times 1.1 = 2.2$ seconds For 50 oscillations: $50 \times 2.2 = 110.0$ seconds. The answer is A) 110.0 seconds. Why the wrong answers occur: B) 115.5 seconds results from incorrectly using the percentage increase directly (100 + 15.5% ≈ 115.5), confusing linear relationships with the square root relationship. C) 121.0 seconds comes from mistakenly applying the 21% increase directly to the time (100 × 1.21), ignoring that period depends on the square root of length. D) 126.5 seconds likely results from compounding errors or misapplying the percentage formula. Study tip: For pendulum problems, always remember that period scales with the square root of length changes, not linearly. Calculate $\sqrt{1 + \text{percentage change}}$ to find the period ratio.
Two identical simple pendulums are set up side by side with slightly different lengths: L₁ = 1.000 m and L₂ = 1.001 m. If they start in phase, after how many oscillations of the shorter pendulum will they first be completely out of phase (180° phase difference)?
Answer: After approximately 1000 oscillations of the shorter pendulum When you encounter coupled oscillator problems involving phase relationships, you're dealing with the phenomenon of beating - where two slightly different frequencies create a periodic pattern of constructive and destructive interference. For simple pendulums, the period is $T = 2\pi\sqrt{\frac{L}{g}}$ . Since the lengths differ slightly, the pendulums have different periods and will gradually shift relative to each other. The shorter pendulum (L₁ = 1.000 m) oscillates faster than the longer one (L₂ = 1.001 m). To find when they're 180° out of phase, you need to determine when the faster pendulum has completed exactly one-half oscillation more than the slower one. The period difference is approximately $\Delta T = T_2 - T_1 \approx T_1 \cdot \frac{\Delta L}{2L_1} = T_1 \cdot \frac{0.001}{2(1.000)} = T_1 \cdot 0.0005$ . For a 180° phase shift, the time difference must equal $\frac{T_1}{2}$ . Setting up the equation: $n \cdot \Delta T = \frac{T_1}{2}$ , where n is the number of oscillations. Solving: $n = \frac{T_1}{2\Delta T} = \frac{1}{2(0.0005)} = 1000$ . Choice A (500 oscillations) represents only a 90° phase shift - they'd be in quadrature, not opposition. Choice C (1500 oscillations) would put them 270° out of phase. Choice D (2000 oscillations) would bring them back in phase (360° difference). Remember: in beating problems, the beat frequency equals the difference in oscillation frequencies, and phase relationships repeat in predictable patterns based on this frequency difference.
Two simple pendulums have the same length but oscillate in different locations. Pendulum A has a period of 2.0 s on Earth's surface, while pendulum B has a period of 2.4 s on the surface of another planet. What is the ratio of gravitational acceleration on the planet to that on Earth?
Answer: $\frac{g_{planet}}{g_{Earth}} = \left(\frac{2.0}{2.4}\right)^2 = 0.69$ When you encounter pendulum problems involving different gravitational fields, focus on the period formula and how gravitational acceleration affects oscillation timing. For a simple pendulum, the period is $T = 2\pi\sqrt{\frac{L}{g}}$ , where L is length and g is gravitational acceleration. Since both pendulums have identical lengths, you can set up a ratio to eliminate the unknown length and solve for the gravitational accelerations. Starting with the period formula for each location: - Earth: $T_E = 2\pi\sqrt{\frac{L}{g_E}} = 2.0 \text{ s}$ - Planet: $T_P = 2\pi\sqrt{\frac{L}{g_P}} = 2.4 \text{ s}$ Taking the ratio $\frac{T_P}{T_E} = \frac{2.4}{2.0} = 1.2$ , this equals $\sqrt{\frac{g_E}{g_P}}$ . Therefore: $1.2 = \sqrt{\frac{g_E}{g_P}}$ Squaring both sides: $1.44 = \frac{g_E}{g_P}$ , so $\frac{g_P}{g_E} = \frac{1}{1.44} = 0.69$ This confirms answer C is correct. A incorrectly uses a direct ratio of periods, ignoring the square root relationship in the formula. B uses the inverse period ratio but still misses the squaring step. D squares the period ratio but uses it in the wrong direction, giving the ratio of Earth's gravity to the planet's gravity instead. Study tip: Remember that period and gravitational acceleration have an inverse square root relationship. When periods increase, gravity decreases, and you must square the period ratio to find the gravity ratio.
A simple pendulum oscillates with amplitude θ₀ = 30°. The restoring torque when the pendulum is at maximum displacement is τ₀. What is the restoring torque when the pendulum passes through an angle of 15°?
Answer: $\tau = \frac{\tau_0}{2}$ When analyzing pendulum motion, remember that the restoring torque is directly proportional to the angular displacement from equilibrium. This relationship comes from the gravitational force component that acts to restore the pendulum to its vertical position. For a simple pendulum, the restoring torque is given by $\tau = -mgL\sin\theta$ , where $m$ is the mass, $g$ is gravitational acceleration, $L$ is the length, and $\theta$ is the angular displacement. Since $mgL$ remains constant for a given pendulum, the torque is directly proportional to $\sin\theta$ . At maximum displacement ( $\theta_0 = 30°$ ), we have $\tau_0 = mgL\sin(30°) = mgL \cdot 0.5$ . At $\theta = 15°$ , the torque becomes $\tau = mgL\sin(15°)$ . Since $\sin(15°) = 0.259$ and $\sin(30°) = 0.5$ , we get $\tau = \frac{\sin(15°)}{\sin(30°)} \cdot \tau_0 = \frac{0.259}{0.5} \cdot \tau_0 \approx \frac{\tau_0}{2}$ . Choice A ( $\tau_0/4$ ) incorrectly assumes a quadratic relationship with angle. Choice C ( $\tau_0/\sqrt{2}$ ) might come from confusing this with energy relationships or trigonometric identities involving 45°. Choice D ( $2\tau_0/3$ ) doesn't correspond to any meaningful physical relationship for these angles. The correct answer is B: $\tau = \frac{\tau_0}{2}$ . Study tip: For pendulum problems, always remember that restoring torque depends on $\sin\theta$ , not $\theta$ itself. This sine relationship is crucial for understanding the non-linear nature of large-amplitude pendulum motion.
Two identical simple pendulums are set up with the same length $L$ . Pendulum A is given an initial angular displacement of $5°$ and released from rest. Pendulum B is given an initial angular displacement of $15°$ and released from rest. Assuming the small angle approximation is valid for pendulum A but not for pendulum B, how do their periods compare?
Answer: Pendulum B has a slightly longer period than pendulum A due to nonlinear restoring force For small angles, $\sin\theta \approx \theta$ and the period is $T = 2\pi\sqrt{L/g}$ , independent of amplitude. However, for larger amplitudes where the small angle approximation fails, $\sin\theta < \theta$ , making the restoring force smaller than the linear approximation predicts. This results in a longer period. The exact period for large amplitudes involves elliptic integrals and increases with amplitude. Choice A ignores the amplitude dependence for large angles. Choice C incorrectly suggests the period decreases. Choice D is wrong because pendulum period is independent of mass for both small and large amplitudes.
A simple pendulum clock keeps accurate time at sea level where g = 9.80 m/s². If the clock is moved to an altitude where g = 9.78 m/s², how much time will the clock lose per day?
Answer: The clock will lose approximately 88 seconds per day When you encounter pendulum problems involving changes in gravitational acceleration, remember that the period of a simple pendulum depends on $T = 2\pi\sqrt{\frac{L}{g}}$ . Since the period is inversely proportional to $\sqrt{g}$ , when gravity decreases, the period increases, making the clock run slower. At sea level, the period is $T_1 = 2\pi\sqrt{\frac{L}{9.80}}$ . At altitude, it becomes $T_2 = 2\pi\sqrt{\frac{L}{9.78}}$ . To find the fractional change in period: $\frac{T_2}{T_1} = \sqrt{\frac{9.80}{9.78}} = 1.001022$ This means each pendulum swing takes about 0.1022% longer than it should. Over 24 hours (86,400 seconds), the clock will lose: $86,400 \times 0.001022 = 88.3$ seconds. Looking at the answer choices: A) underestimates the time loss by roughly half, likely from an error in the calculation or using an approximation that's too crude. C) doubles the correct answer, possibly from confusing the fractional change formula or making an algebraic error. D) incorrectly suggests the clock gains time, which would only happen if gravity increased rather than decreased. The correct answer is B) approximately 88 seconds per day. Study tip: For pendulum problems, always remember that decreasing gravity means longer periods and slower clocks. Use the relationship $\frac{\Delta T}{T} \approx -\frac{1}{2}\frac{\Delta g}{g}$ for small changes in gravity as a quick check on your calculations.
A uniform solid cylinder of radius R and mass M is suspended as a physical pendulum by a knife edge tangent to its curved surface. What is the ratio of this pendulum's period to that of a simple pendulum of length R?
Answer: $\frac{T_{cylinder}}{T_{simple}} = \sqrt{\frac{3}{2}}$ When you encounter a physical pendulum problem, you need to apply the formula for the period of a physical pendulum and compare it to a simple pendulum. The key is correctly identifying the moment of inertia and the distance from the pivot to the center of mass. For a physical pendulum, the period is $T = 2\pi\sqrt{\frac{I}{mgd}}$ , where $I$ is the moment of inertia about the pivot point, $m$ is mass, and $d$ is the distance from pivot to center of mass. For the cylinder suspended at its edge, $d = R$ (distance from edge to center). The moment of inertia about the edge uses the parallel axis theorem: $I = I_{center} + md^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$ . Therefore: $T_{cylinder} = 2\pi\sqrt{\frac{3MR^2/2}{MgR}} = 2\pi\sqrt{\frac{3R}{2g}}$ For a simple pendulum of length $R$ : $T_{simple} = 2\pi\sqrt{\frac{R}{g}}$ Taking the ratio: $\frac{T_{cylinder}}{T_{simple}} = \sqrt{\frac{3R/2g}{R/g}} = \sqrt{\frac{3}{2}}$ This confirms answer B is correct. Answer A ( $\sqrt{2/3}$ ) inverts this ratio. Answer C ( $\sqrt{4/3}$ ) likely comes from incorrectly using $I = MR^2$ instead of applying the parallel axis theorem. Answer D ( $\sqrt{3/4}$ ) combines both errors—using wrong moment of inertia and inverting the ratio. Remember: physical pendulum problems always require the parallel axis theorem when the pivot isn't at the center of mass, and the period is typically longer than the equivalent simple pendulum.

Start this quiz