Question
Word Problems: Remember to use GRASP and show string number before stating final answer. 1) Abobon the end of a spring is pulled down and released. If the bob moves up and down 12 times in 3.6 seconds, determine the period and frequency of the oscillation [Ans: 0.30s;3.3Hz] 2) A piston in a car moves up a down 150 times per minute (150 rpm)Find the frequency in Hz and the period of the vibration. [Ans: 2.5Hz;0.40s] 3) The frequency of a note produced by a tuning fork is 440 Hz. Find the wavelength of the sound given a sound speed of (a) 332m/s (b) 344m/s. [Ans: 0.755 m 0.782 m] 4) The frequency of another tuning fork is 512 Hz. If the wavelength of the vibration is 66.4 cm, find the speed of sound (Ans: 3.40times 10^2m/s) 5) Determine the speed of a wave with a frequency of 1025 MHz and a wavelength of 2.9 m (Ans: 3.0times 10^8m/s]
Solution
4.4
(115 Votes)
Amelie
Master · Tutor for 5 years
Answer
## Word Problem Solutions:**1) Oscillating Bob*** **Given:** 12 oscillations in 3.6 seconds* **Required:** Period (T) and Frequency (f)* **Relationship:** f = 1/T and T = 1/f* **Solution:** * T = Total time / Number of oscillations * T = 3.6 s / 12 oscillations * T = 0.30 s * f = 1 / T * f = 1 / 0.30 s * f = 3.3 Hz* **Answer 1:** Period (T) = 0.30 s; Frequency (f) = 3.3 Hz**2) Car Piston*** **Given:** 150 oscillations per minute (rpm)* **Required:** Frequency (f) in Hz and Period (T)* **Relationship:** f = oscillations/time and T = 1/f* **Solution:** * Convert rpm to Hz: * f = 150 oscillations / 60 seconds * f = 2.5 Hz * T = 1/f * T = 1 / 2.5 Hz * T = 0.40 s* **Answer 2:** Frequency (f) = 2.5 Hz; Period (T) = 0.40 s**3) Tuning Fork (440 Hz)*** **Given:** Frequency (f) = 440 Hz; Speed of sound (v) = (a) 332 m/s, (b) 344 m/s* **Required:** Wavelength (λ)* **Relationship:** v = fλ* **Solution (a):** * λ = v / f * λ = 332 m/s / 440 Hz * λ = 0.755 m* **Solution (b):** * λ = v / f * λ = 344 m/s / 440 Hz * λ = 0.782 m* **Answer 3:** Wavelength (λ) = (a) 0.755 m; (b) 0.782 m**4) Tuning Fork (512 Hz)*** **Given:** Frequency (f) = 512 Hz; Wavelength (λ) = 66.4 cm = 0.664 m* **Required:** Speed of sound (v)* **Relationship:** v = fλ* **Solution:** * v = fλ * v = 512 Hz * 0.664 m * v = 340 m/s* **Answer 4:** Speed of sound (v) = 3.40 x 10² m/s**5) Wave Speed*** **Given:** Frequency (f) = 1025 MHz = 1.025 x 10⁹ Hz; Wavelength (λ) = 2.9 m* **Required:** Speed of wave (v)* **Relationship:** v = fλ* **Solution:** * v = fλ * v = (1.025 x 10⁹ Hz) * 2.9 m * v = 2.97 x 10⁹ m/s (approximately 3.0 x 10⁸ m/s given potential rounding in the problem)* **Answer 5:** Speed of wave (v) = 3.0 x 10⁸ m/s (approximated)