Question

A block with a speaker bolted to it is connected to a spring having spring constant k = 24.0 N/m and oscillates as shown in the figure below. The total mass of the block and speaker is 3.00 kg, and the amplitude of this unit's motion is 0.500 m. The speaker emits sound waves of frequency 500 Hz. www (a) Determine the highest frequencies heard by the person to the right of the speaker. (Give your answer correct to 1 Hz.) (b) Determine the lowest frequencies heard by the person to the right of the speaker. (Give your answer correct to 1 Hz.) Hz (c) If the maximum sound level heard by the person is 69.0 dB when the speaker is at its closest distance d - 1.00 m from him, what is the minimum sound level heard by the observer? dB

Can you please explain all parts

I meant to ask this question

Answer 1

Answer:

(a) 502

(b) 498

(c) 63

For tuning fork

Answer: 8.10

wwwwww Giveni K=24.0 N/m ms 3.00 kg A=0.500 m Complitude of m). treguney of sound – 500 m2 with mam speed of im, men wika mer - sumt - D x 8.5 m - 8 x = V2 m/s speed of sound ein air - Bages 343 m/s Now, 6) Frequency in highest if source is coming towards observer wil mandman speed f ) - 348 – 502 He Fresnenng to bewert it sauce is going away from wenye vy mbo 349 Ve observed with SE mys - A 500 Me 0 349+

Š x soo Me 328+ V2 N 498 He sondness (8) = 10000 to B,- B = lo logo also, ademaitze 2 (distance + -he relay, here maximum sound seid =P soy. - minimum sand terel=2=2 closest ditance = 100m d farthest distance = dtim = 2.00m Amplitude = 0.500m a distance men=2.

= 69.008 og 20 kg. (5) = 69.0dB – 20t09 (2) To 69.0db – 6.02 dB = 62.9794 dB 63 dB A tuning fork vibrating at 508 H2 = to (ay) released from rest with donward acce" – 9.8 m/s2 и со Cef=.490 Hz fa Gav) 4 (variety IV = sound speed in air V = 9.8 m/s2

- Vev, so 490 430 = 343 343 343 tve - 1508 8808 - 342+v. 355.6. V = 12.6 NOW, = 42 +224 1 - 0 1 2 3 4.8 VÀ h . 8. 1 m.