Q1: What is the wavelength of a radio wave broadcasted by a radio station with a frequency of 1300 kHz? Where the speed of the radio wave is 3 × 10⁸ m/s.
Given:
Frequency, f = 1300 kHz = 1300 × 10³ Hz = 1.3 × 10⁶ Hz
Speed, v = 3 × 10⁸ m/s
Formula: v = fλ
λ = v / f
Calculation:
λ = (3 × 10⁸) / (1.3 × 10⁶)
λ = (3 × 10⁸) / (1.3 × 10⁶)
λ = (3/1.3) × 10²
λ = 2.3076 × 10²
λ = 230.76 m
Answer: 230.76 m
Frequency, f = 1300 kHz = 1300 × 10³ Hz = 1.3 × 10⁶ Hz
Speed, v = 3 × 10⁸ m/s
Formula: v = fλ
λ = v / f
Calculation:
λ = (3 × 10⁸) / (1.3 × 10⁶)
λ = (3 × 10⁸) / (1.3 × 10⁶)
λ = (3/1.3) × 10²
λ = 2.3076 × 10²
λ = 230.76 m
Answer: 230.76 m
Q2: The waves moving in the pond have a wavelength of 1.6 m, and a frequency of 0.80 Hz. Calculate the speed of these water waves.
Given:
Wavelength, λ = 1.6 m
Frequency, f = 0.80 Hz
Formula: v = fλ
Calculation:
v = 0.80 × 1.6
v = 1.28 m/s
Answer: 1.28 m/s
Wavelength, λ = 1.6 m
Frequency, f = 0.80 Hz
Formula: v = fλ
Calculation:
v = 0.80 × 1.6
v = 1.28 m/s
Answer: 1.28 m/s
Q3: If 50 waves pass through a point in the rope in 10 seconds, what are the frequency and the period of the wave? If its wavelength is 8 cm, calculate the wave speed.
Given:
Number of waves = 50
Time, t = 10 s
Wavelength, λ = 8 cm = 0.08 m
Formulas:
Frequency, f = Number of waves / Time
Period, T = 1/f
Wave speed, v = fλ
Calculation:
f = 50 / 10 = 5 Hz
T = 1/5 = 0.2 s
v = 5 × 0.08 = 0.4 m/s
Answer: Frequency = 5 Hz, Period = 0.2 s, Wave speed = 0.4 m/s
Number of waves = 50
Time, t = 10 s
Wavelength, λ = 8 cm = 0.08 m
Formulas:
Frequency, f = Number of waves / Time
Period, T = 1/f
Wave speed, v = fλ
Calculation:
f = 50 / 10 = 5 Hz
T = 1/5 = 0.2 s
v = 5 × 0.08 = 0.4 m/s
Answer: Frequency = 5 Hz, Period = 0.2 s, Wave speed = 0.4 m/s
Q4: A slinky has produced a longitudinal wave. The wave travels at a speed of 40 cm/s and the frequency of the wave is 20 Hz. What is the minimum separation between the consecutive compressions?
Given:
Speed, v = 40 cm/s = 0.4 m/s
Frequency, f = 20 Hz
Formula: v = fλ
λ = v / f
Calculation:
λ = 0.4 / 20
λ = 0.02 m
Answer: 0.02 m (minimum separation between compressions is the wavelength)
Speed, v = 40 cm/s = 0.4 m/s
Frequency, f = 20 Hz
Formula: v = fλ
λ = v / f
Calculation:
λ = 0.4 / 20
λ = 0.02 m
Answer: 0.02 m (minimum separation between compressions is the wavelength)
Q5: Suppose a student is generating waves in a slinky. The student's hand makes one complete orbit and back oscillation in 0.40 s. The wavelength in the slinky is 0.60 m. For this wave, determine a. Period and frequency b. Wave speed
Given:
Time for one complete oscillation, T = 0.40 s
Wavelength, λ = 0.60 m
Formulas:
Frequency, f = 1/T
Wave speed, v = fλ
Calculation (a):
Period, T = 0.40 s
Frequency, f = 1/0.40 = 2.5 Hz
Calculation (b):
v = 2.5 × 0.60
v = 1.5 m/s
Answer: a) Period = 0.40 s, Frequency = 2.5 Hz
b) Wave speed = 1.5 m/s
Time for one complete oscillation, T = 0.40 s
Wavelength, λ = 0.60 m
Formulas:
Frequency, f = 1/T
Wave speed, v = fλ
Calculation (a):
Period, T = 0.40 s
Frequency, f = 1/0.40 = 2.5 Hz
Calculation (b):
v = 2.5 × 0.60
v = 1.5 m/s
Answer: a) Period = 0.40 s, Frequency = 2.5 Hz
b) Wave speed = 1.5 m/s
Q6: If 80 compressions pass through a point in spring in 20 seconds. Calculate the frequency and the period? If two consecutive compressions are 8 cm apart, calculate the wave speed.
Given:
Number of compressions = 80
Time, t = 20 s
Distance between compressions (wavelength), λ = 8 cm = 0.08 m
Formulas:
Frequency, f = Number of compressions / Time
Period, T = 1/f
Wave speed, v = fλ
Calculation:
f = 80 / 20 = 4 Hz
T = 1/4 = 0.25 s
v = 4 × 0.08 = 0.32 m/s
Answer: Frequency = 4 Hz, Period = 0.25 s, Wave speed = 0.32 m/s
Number of compressions = 80
Time, t = 20 s
Distance between compressions (wavelength), λ = 8 cm = 0.08 m
Formulas:
Frequency, f = Number of compressions / Time
Period, T = 1/f
Wave speed, v = fλ
Calculation:
f = 80 / 20 = 4 Hz
T = 1/4 = 0.25 s
v = 4 × 0.08 = 0.32 m/s
Answer: Frequency = 4 Hz, Period = 0.25 s, Wave speed = 0.32 m/s
Q7: Waves on a swimming pool propagate at 0.90 m/s. If you splash the water at one end of the pool, observe the wave to go to the opposite end, reflect, and return in 30.0 s. How far away is the other end of the pool?
Given:
Speed, v = 0.90 m/s
Total time for round trip, t = 30.0 s
Formula:
Distance for round trip = v × t
Distance to one end = (v × t) / 2
Calculation:
Round trip distance = 0.90 × 30 = 27 m
Distance to opposite end = 27 / 2 = 13.5 m
Answer: 13.5 m
Speed, v = 0.90 m/s
Total time for round trip, t = 30.0 s
Formula:
Distance for round trip = v × t
Distance to one end = (v × t) / 2
Calculation:
Round trip distance = 0.90 × 30 = 27 m
Distance to opposite end = 27 / 2 = 13.5 m
Answer: 13.5 m
Q8: A simple oscillating pendulum has a length of 80.0 cm. Calculate its a. Period b. Frequency. When g = 9.8 m/s²
Given:
Length, L = 80.0 cm = 0.80 m
g = 9.8 m/s²
Formulas:
Period, T = 2π √(L/g)
Frequency, f = 1/T
Calculation (a):
T = 2 × 3.14 × √(0.80/9.8)
T = 6.28 × √(0.08163)
T = 6.28 × 0.2857
T = 1.794 s
Calculation (b):
f = 1 / 1.794
f = 0.557 Hz
Answer: a) Period = 1.794 s, b) Frequency = 0.557 Hz
Length, L = 80.0 cm = 0.80 m
g = 9.8 m/s²
Formulas:
Period, T = 2π √(L/g)
Frequency, f = 1/T
Calculation (a):
T = 2 × 3.14 × √(0.80/9.8)
T = 6.28 × √(0.08163)
T = 6.28 × 0.2857
T = 1.794 s
Calculation (b):
f = 1 / 1.794
f = 0.557 Hz
Answer: a) Period = 1.794 s, b) Frequency = 0.557 Hz
