A spring-mass oscillator has spring constant k and mass m. If you replace the spring with one that is four times stiffer (4k) while keeping the same mass, what happens to the period?
AThe period doubles — a stiffer spring has more force, so it takes longer to complete a cycle
BThe period halves — the system oscillates twice as fast
CThe period decreases by a factor of √2
DThe period is unchanged — it depends only on mass
T = 2π√(m/k). With k replaced by 4k, T becomes 2π√(m/4k) = 2π·(1/2)√(m/k) = T/2. The period halves. The common misconception is that a stiffer spring should move more slowly because it exerts a larger force — but larger k means larger restoring force at every displacement, which produces faster acceleration back to equilibrium and therefore faster oscillation. Stiffer springs always decrease period.
Question 2 Multiple Choice
A student pulls a mass on a spring to amplitude A and releases it, timing the period. Then they pull the same mass to amplitude 2A and release it again. What do they measure for the second period?
ADouble the first period — a larger amplitude requires more distance to travel
B√2 times the first period — energy is proportional to A², so velocity scales as √2
CThe same period as the first — period is amplitude-independent in SHM
DHalf the first period — more energy means faster oscillation
Period T = 2π√(m/k) depends only on k and m, not on amplitude. Doubling amplitude doubles the distance traveled in a cycle AND doubles the maximum velocity (v_max = Aω), so both effects cancel and the period is unchanged. This amplitude-independence is a unique and important property of SHM. The energy stored (½kA²) increases with amplitude, but energy determines how vigorously the system oscillates, not how fast.
Question 3 True / False
The total mechanical energy of a spring-mass system equals ½kA² at every point in the oscillation cycle, including the equilibrium position.
TTrue
FFalse
Answer: True
By conservation of energy, PE + KE = constant = ½kA² throughout the cycle. At equilibrium (x = 0), PE = 0 and KE = ½mv_max² = ½kA². At maximum displacement (x = A), KE = 0 and PE = ½kA². At any intermediate point, the two forms trade off but the sum is always ½kA². The total energy is set by the amplitude; the frequency is set by k and m independently.
Question 4 True / False
A stiffer spring (larger k) results in a longer period because the larger restoring force means the mass takes more time to complete each oscillation.
TTrue
FFalse
Answer: False
This intuition is backwards. A stiffer spring exerts a larger restoring force at every displacement, so the acceleration back toward equilibrium is larger, and the system oscillates faster. T = 2π√(m/k): since k is in the denominator under the square root, increasing k decreases T. The confusion arises from conflating 'larger force' with 'slower motion' — in SHM, the larger force produces larger accelerations and therefore faster oscillation.
Question 5 Short Answer
Why can you change the amplitude of a spring-mass oscillator without affecting its frequency of oscillation?
Think about your answer, then reveal below.
Model answer: Frequency depends only on k and m (ω = √(k/m)), not on how far the mass is displaced. Amplitude sets the energy stored (E = ½kA²) and the maximum speed (v_max = Aω), but both scale together — the extra distance of a larger amplitude is traversed at proportionally higher speeds. Physically, a larger amplitude means larger restoring forces AND larger velocities at every point, so the two effects cancel and the time per cycle is unchanged.
This amplitude-independence comes directly from Hooke's law being linear. Because F = −kx grows proportionally with displacement, doubling amplitude doubles both the force (and hence acceleration) and the distance, leaving the time to traverse the cycle unchanged. Nonlinear restoring forces break this symmetry, causing period to depend on amplitude — which is why large-angle pendulums drift and why this approximation is only valid for small displacements.