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1. Herr Gardner’s radio show is on 91.7 MHz every Saturday night. What is the wavelength of these waves? v =fλ 3 108 m/s = (91.7 106Hz)(λ) × × 300,000,000 λ= meters = 3.27meters 91,700,000 2. A guitar string has a length of 1.20 meters with a fundamental frequency of 250 Hz and a linear density of 5.5 10−4 kg/m. What is the tension of the string? × tension(N) v = string slinear density(kg/m) v =fλ string tension(N) therefore =fλ slinear density(kg/m) tension =250Hz (λ) 5.5 10−4kg/m r × tension =250(λ) 5.5 10−4 r × √tension=√5.5 10−4(250)(λ) × √tension=5.863λ tension=34.375λ2 1 We know that ℓ = λ. We can rearrange to get λ=2ℓ , so λ=2.4m string string 2 Plugging that back in we get tension=34.375(2.4)2, and tension= 198N 3. What is the length of a tube open at both ends with the same fundamental frequency as the guitar string from the previous problem? Assume the speed of sound is 340m/s. v =fλ 340m/s=250Hz(λ) 1 ℓ= λ, so λ=2ℓ 2 Plugging in λ, we get 340=250(2ℓ), so ℓ=0.68m 4. A tuning fork of an unknown frequency produces 6 beats per second when sounded with a 440Hz tuning fork. The beat frequency is greater when the unknown fork is sounded with a 430 Hz tuning fork. What is the frequency of the unknown fork? Remember that beats per second is calculated with the expression f1 f2 , where | − | f1 and f2 are the two frequencies vibrating simultaneously. We know f 440Hz = 6Hz, so f = 434Hz or 446Hz. unknown unknown | − | But since the beat frequency is greater with the 430Hz fork, then f must be 446Hz unknown FORMULAS: The Wave Equation: v =fλ Velocity of Sound travelling through a string under tension(T): T v = slinear density Frequency of any harmonic in a closed tube of length ℓ: v =fλ n 4ℓ ℓ= λ, so λ= 4 n 4ℓ nv v =f , therefore f = n 4ℓ Frequency of any harmonic in an open tube of length ℓ: v =fλ n 2ℓ ℓ= λ, so λ= 2 n 2ℓ nv v =f , therefore f = n 2ℓ Length of a string vibration at the third overtone: n 3 ℓ= λ, so ℓ= λ 2 2 5. The tension on the 1.6 meter long wire is 135.0 N. The wire has a total mass of 0.008kg. What is the frequency of the THIRD OVERTONE produced by the wire? 0.008kg Linear Density: = 0.005kg/m 1.6m tension =fλ linear density r 135N =fλ 0.005kg/m r 2 n 3 2 ℓ= λ, so for the third harmonic, ℓ= λ, so λ= ℓ 2 2 3 135N 2 Plugging this in, we get =f( )(1.6m). 0.005kg/m 3 r 164.317=1.067f,therefore, f =154.05Hz 6. What is the wavelength of the next harmonic produced by the 1.6 meter wire shown above? n We know that ℓ= λ. Knowing that this is the fourth harmonic, the equation becomes ℓ=2λ. 2 ℓ=1.6m, so λ=0.8m . 7. What is the fundamental frequency of a 0.8 meter closed pipe? n 1 We know that for a closed pipe, ℓ= λ, so for the fundamental frequency, ℓ= λ. 4 4 Plugging in ℓ as 0.8m, we get λ=3.2m. v =fλ 340m/s=f(3.2m), so f =106.25Hz 8. 9. What is the fundamental frequency of a 0.8 meter open pipe? n ℓ= λ, so λ=2ℓ. Substituting, we get λ=2(0.8m)=1.6m 2 v =fλ 340m/s=f(1.6m) f = 212.5Hz 10. 11.Write the electromagnetic spectrum starting from the longest wavelength. Radio Waves Microwaves Infared Visible Light UV X-Rays Gamma Rays −→ −→ −→ −→ −→ −→ Visible light: Red Orange Yellow Green Blue Violet −→ −→ −→ −→ −→ 3