1) The arrangement in the drawing shows a block (mass = 15.0 kg) that is held in

Question

1) The arrangement in the drawing shows a block (mass = 15.0 kg) that is held in position on a frictionless incline by a cord (length = 0.584 m). The mass per unit length of the cord is 1.10 × 10-2 kg/m, so the mass of the cord is negligible compared to the mass of the block. The cord is being vibrated at a frequency of 168 Hz (vibration source not shown in the drawing). What is the smallest angle between 15.0 ° and 90.0 ° at which a standing wave exists on the cord?

2) A row of seats is parallel to a stage at a distance of 8.5 m from it. At the center and front of the stage is a diffraction horn loudspeaker. This speaker sends out its sound through an opening that is like a small doorway with a width D of 0.078 m. The speaker is playing a tone that has a frequency of 5.00 × 104 Hz. The speed of sound is 343 m/s. What is the separation between two seats, located near the center of the row, at which the tone cannot be heard?

Explanation / Answer

1) there are several elements to bring together in this question...

first, the tension in the cord must equal the component of the block's weight down the plane; this tension will equal

T = m g sin(theta) where m is the mass of the block

the tension enters this problem since the speed of sound on a string is related to the tension according to:

c = Sqrt[T/u] where T is the tension and u is the mass density of the string (given as 0.0110 kg/m)

and the speed is important since the frequency of the resonances on a string is dependent on the speed of sound in the string

so, what are the allowable standing wave frequencies?

for a string clamped at both ends, the fundamental frequency has a wavelength that is 2 the length of the wire (in other words, the length of the string is one half a wavelength of the fundamental mode of vibration), so we can write the frequency of the fundamental mode as

f = c/2L where L is the length of the string

the allowable frequencies are integral multiples of the fundamental, so we have

f = n c/2L or c = 2 L f/n

we know L and f, but don't know n....let's see how we figure out which harmonic satisfies the requirement for the smallest angle greater than 15 degrees to have a standing wave....


we know that c = Sqrt[T/u] and that T = mg sin(theta)

substitute this expression for T and c = 2Lf/n and get

2 L f/n = Sqrt[ m g sin(theta)/u]

square both sides and solve for sin(theta)

sin(theta) = 4 L^2 f^2 u/(m g n^2)

setting L=0.584m, f = 168Hz, u = 0.0110kg/m, m = 15.0kg and g = 9.81 m/s/s, we get

sin(theta) = 4 L^2 f^2 u/(m g n^2)

= 4*0.584^2*168^2*0.0110 /15*9.81*n^2

2.88/n^2 = sin(theta)

now, we know that sin(theta) can only have values between 1 and -1, and that n is an integer, so we can see immediately that n=1 is not a solution for this equation...so...let's see what we get for n = 2, n=3, etc

if n=2, then sin(theta) = 2.88/2^2 => theta =46.05 deg .... within the range, but is it the smallest value?

if n=3, sin(theta) = 2.88/3^2 => theta =18.66 deg....let's do n =4:

if n=4, sin(theta) =2.88/4^2 => theta = 10.36 deg...

so, we have our answer, the smallest angle (greater than 15 deg) occurs for n=3, and that angle is 18.66 deg

2)

For a single slit diffraction the dark fringes are located by
sin() = m/a where m = 1; c/f = 343/5.0x10^4 = 0.00686m ; a = 0.078m

So = arcsin(0.00686/0.078) = 0.0879 =5.04degree

So the separation distance = 2*8.5m*tan(5.04) = 1.49m

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