The equation of a sound wave in air is given by Δp = ( 0 . 02 ) sin [ ( 3000 ) t - ( 9 . 0 ) x ] , where all variables are in SI units. If the equilibrium pressure of air is 1 . 01 × 10 5 N / m 2 , what are the maximum and minimum pressures at a point as the wave passes through that point?

If we look at the equation Δp = ( 0 . 02 ) sin [ ( 3000 ) t - ( 9 . 0 ) x ] , we can say that the pressure amplitude is Δp max = 0 . 02 N / m 2 The equilibrium pressure given to us is P e q = 1 . 01 × 10 5 N / m 2 . The maximum and minimum pressure is given by: P e q ± △ p m a x Hence, the maximum and minimum pressures at a point in the wave motion will be 1 . 01 × 10 5 ± 0 . 02 N / m 2 .

The equation of a sound wave in air is given by Δp = ( 0 . 02 ) sin [ ( 3000 ) t - ( 9 . 0 ) x ] , where all variables are in SI units. If the equilibrium pressure of air is 1 . 01 × 10 5 N / m 2 , what are the maximum and minimum pressures at a point as the wave passes through that point?

If we look at the equation Δp = ( 0 . 02 ) sin [ ( 3000 ) t - ( 9 . 0 ) x ] , we can say that the pressure amplitude is Δp max = 0 . 02 N / m 2 The equilibrium pressure given to us is P e q = 1 . 01 × 10 5 N / m 2 . The maximum and minimum pressure is given by: P e q ± △ p m a x Hence, the maximum and minimum pressures at a point in the wave motion will be 1 . 01 × 10 5 ± 0 . 02 N / m 2 .

Determine the change in volume of 6 liters of alcohol if the pressure is decreased from 200 cm of Hg to 75 cm . [velocity of sound in alcohol is 1280 m / s , density of alcohol = 0 . 81 g / cc , density of Hg = 13 . 6 gm / cc and g = 9 . 81 m / s

The velocity of sound in liquid v = B ρ ⇒ B = v 2 ρ We also know that B = - ΔP ( ΔV / V ) So, we can say that: - V Δ P Δ V = v 2 ρ ⇒ ΔV = V ( - ΔP ) ρv 2 Here Δ P = H 2 ρ g - H 1 ρ g ⇒ Δ P = ( 75 - 200 ) × 13 . 6 × 981 = - 1 . 667 × 10 6 dynes / cm 2 Now, if we substitute the values, we will get: Δ V = 6 × 10 3 1 . 667 × 10 6 0 . 81 × 1 . 280 × 10 5 2 = 0 . 75 cc Hence, the change in volume is 0 . 75 c c

Calculate the stress in a tight wire of a material whose Young's modulus is 19 . 6 × 10 11 dyne/cm 2 so that the speed of the longitudinal waves is 10 times the speed of transverse waves.

The speed of transverse waves in the string is given by the relation: c 1 = T m = T A d Here, A is the cross-sectional area and d is the density. The speed of longitudinal waves in string is c 2 = Y / d . Also, we have, c 2 = 10 c 1 ⇒ Y d = 10 T A d ⇒ Stress = T A = Y 100 = 19 . 6 × 10 9 dyne / cm 2 Hence, the stress in the wire is 19 . 6 × 10 9 dyne / cm 2