Question

A tuning fork on a movable cart is vibrating at a frequency of 1500 Hz. As you are walking past the cart at 2 m/sec, the cart begins accelerating uniformly in the direction you’re walking. When it is 20 m in front of you, the sound of the tuning fork is heard to be at a frequency of 1380 Hz. Knowing that the air temperature at your location is 23.6°C, determine the acceleration of the cart.

Answer 1

speed of sound at any temperature (T) = 331 + (0.61)(T)

at T = 23.6 degree

speed of sound (v) = 331 + (0.61*23.6)

v = 345.4 m/s

now , original emitted frequency = 1500 Hz

observed frequency = 1380 Hz

Using, doppler's effect of sound

F_observed = F_emitted (v + V_observer)/(v-V_source)

1380 = 1500 (345.4 +2)/(345.4 + V_cart)

345.4 + V_cart = 377.61

V_cart = 32.21 m/s

Now, let the distance travelled by man = x

x = 2 * t   ..(1)

distance travelled by cart = x + 20

v^2 - u^2 = 2ad

32.21^2 - 0 = 2 * a * (x + 20 )

518.74 = a(x +20)  

518.74 = ax +20a     ..(2)

also, v = u + at

32.21 = 0 + a(t )

put value from (1)

32.21 = a (x/2)

64.42 = ax

put value in equation (2)

518.74 = 64.42 + 20a

454.32 = 20a

a = 22.72 m/s^2

Therefore, acceleration of the cart = 22.72 m/s^2

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