Frequency and Period of SHM - AP Physics 1

  1. For a mass-spring system, what is the effect of increasing spring constant on the period?
    Decreases the period. Period is inversely proportional to square root of spring constant.
  2. What is the frequency of a simple pendulum with period 5 s5 \text{ s}?
    f=0.2 Hzf = 0.2 \text{ Hz}. Using f=1T=15=0.2f = \frac{1}{T} = \frac{1}{5} = 0.2 Hz.
  3. Calculate the period for a spring system with k=200 N/mk = 200 \text{ N/m} and m=0.5 kgm = 0.5 \text{ kg}.
    T=2π0.5200≈0.31 sT = 2\pi \sqrt{\frac{0.5}{200}} \approx 0.31 \text{ s}. Using the spring-mass formula with given values.
  4. What is the period of a wave with a frequency of 20 Hz20 \text{ Hz}?
    T=0.05 sT = 0.05 \text{ s}. Using T=1f=120=0.05T = \frac{1}{f} = \frac{1}{20} = 0.05 s.
  5. What is the formula for the period of a mass-spring system?
    T=2πmkT = 2\pi \sqrt{\frac{m}{k}}. Period depends on mass and spring constant for elastic oscillations.
  6. Calculate the period for a simple pendulum on the Moon with L=1 mL = 1 \text{ m} and g=1.6 m/s2g = 1.6 \text{ m/s}^2.
    T=2π11.6≈4.97 sT = 2\pi \sqrt{\frac{1}{1.6}} \approx 4.97 \text{ s}. Using the pendulum formula with Moon's gravity.
  7. What happens to the angular frequency if the period doubles?
    Halves. Angular frequency is inversely proportional to period.
  8. What determines the period of a simple harmonic oscillator?
    Mass and spring constant. These are the only factors in the period formula for springs.
  9. Calculate the angular frequency for a system with frequency 8 Hz8 \text{ Hz}.
    ω=16π rad/s\omega = 16\pi \text{ rad/s}. Using ω=2πf=2π(8)=16π\omega = 2\pi f = 2\pi(8) = 16\pi rad/s.
  10. If a mass-spring system oscillates at 10 Hz10 \text{ Hz}, what is its period?
    T=0.1 sT = 0.1 \text{ s}. Using T=1f=110=0.1T = \frac{1}{f} = \frac{1}{10} = 0.1 s.
  11. What is the formula for angular frequency?
    ω=2πf=2πT\omega = 2\pi f = \frac{2\pi}{T}. Angular frequency relates to linear frequency and period.
  12. Calculate the frequency of a pendulum with period 4 s4 \text{ s}.
    f=0.25 Hzf = 0.25 \text{ Hz}. Using f=1T=14=0.25f = \frac{1}{T} = \frac{1}{4} = 0.25 Hz.
  13. For a spring-mass system, if m=9 kgm = 9 \text{ kg} and k=36 N/mk = 36 \text{ N/m}, find TT.
    T=2π936≈3.14 sT = 2\pi \sqrt{\frac{9}{36}} \approx 3.14 \text{ s}. Using the spring-mass formula with given values.
  14. Which factors affect the period of a simple pendulum?
    Length and gravity. These are the only factors in the period formula for pendulums.
  15. What is the angular frequency of a system with T=2 sT = 2 \text{ s}?
    ω=π rad/s\omega = \pi \text{ rad/s}. Using ω=2πT=2π2=π\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi rad/s.
  16. Identify the unit of period in the International System of Units.
    Seconds (s). Standard SI unit for time duration of one complete cycle.
  17. What is the frequency of SHM when the system completes 5050 cycles in 10 s10 \text{ s}?
    f=5 Hzf = 5 \text{ Hz}. Using f=cyclestime=5010=5f = \frac{\text{cycles}}{\text{time}} = \frac{50}{10} = 5 Hz.
  18. What is the formula for the period of a simple pendulum?
    T=2πLgT = 2\pi \sqrt{\frac{L}{g}}. Period depends on length and gravity for small-angle oscillations.
  19. Identify the phase constant in the equation x(t)=Acos⁡(ωt+ϕ)x(t) = A \cos(\omega t + \phi).
    ϕ\phi. The phase constant determines initial conditions.
  20. State the effect of increasing gravity on the period of a pendulum.
    Decreases the period. Period is inversely proportional to square root of gravity.
  21. Find the frequency if the angular frequency is 6π rad/s6\pi \text{ rad/s}.
    f=3 Hzf = 3 \text{ Hz}. Using f=ω2π=6π2π=3f = \frac{\omega}{2\pi} = \frac{6\pi}{2\pi} = 3 Hz.
  22. Identify the unit of frequency in the International System of Units.
    Hertz (Hz). Standard SI unit for cycles per second.
  23. What is the equation for the position of a mass in SHM as a function of time?
    x(t)=Acos⁡(ωt+ϕ)x(t) = A \cos(\omega t + \phi). General form for position in simple harmonic motion.
  24. What is the linear frequency for a wave with angular frequency ω=4π\omega = 4\pi?
    f=2 Hzf = 2 \text{ Hz}. Using f=ω2π=4π2π=2f = \frac{\omega}{2\pi} = \frac{4\pi}{2\pi} = 2 Hz.
  25. Which physical quantity is inversely proportional to period?
    Frequency. These quantities are reciprocals of each other.
  26. For a mass-spring system, what is the effect of increasing mass on the period?
    Increases the period. Period is proportional to square root of mass.
  27. Find the frequency if the period is 0.5 s0.5 \text{ s}.
    f=2 Hzf = 2 \text{ Hz}. Using f=1T=10.5=2f = \frac{1}{T} = \frac{1}{0.5} = 2 Hz.
  28. What is the period of oscillation for a spring with k=100 N/mk = 100 \text{ N/m} and m=4 kgm = 4 \text{ kg}?
    T=2π4100≈1.26 sT = 2\pi \sqrt{\frac{4}{100}} \approx 1.26 \text{ s}. Using the spring-mass formula with given values.
  29. State the formula for frequency in terms of angular frequency.
    f=ω2πf = \frac{\omega}{2\pi}. Derived from ω=2πf\omega = 2\pi f, solving for frequency.
  30. How does the period of SHM change if the amplitude is doubled?
    No change. Period is independent of amplitude in SHM.
  31. Find the period if the frequency is 5 Hz5 \text{ Hz}.
    T=0.2 sT = 0.2 \text{ s}. Using T=1f=15=0.2T = \frac{1}{f} = \frac{1}{5} = 0.2 s.
  32. What is the relationship between period and length for a pendulum?
    Directly proportional to L\sqrt{L}. Period increases with square root of length.
  33. What is the dimension of period?
    T. Period has units of time duration.
  34. If a pendulum's length is quadrupled, how does its period change?
    Doubles. Period scales with square root of length, so 4=2\sqrt{4} = 2.
  35. What is the dimension of frequency?
    T⁻¹. Frequency has units of inverse time.
  36. What is the period of a pendulum with L=2 mL = 2 \text{ m} on Mars where g=3.7 m/s2g = 3.7 \text{ m/s}^2?
    T=2π23.7≈4.6 sT = 2\pi \sqrt{\frac{2}{3.7}} \approx 4.6 \text{ s}. Using the pendulum formula with Mars gravity.
  37. What is the period of a pendulum with length 1 m1 \text{ m} on Earth?
    T=2π19.8≈2.01 sT = 2\pi \sqrt{\frac{1}{9.8}} \approx 2.01 \text{ s}. Using the pendulum formula with L=1L = 1 m and g=9.8g = 9.8 m/s².

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