Answer:
Explanation:
b ) The problem is based on Doppler's effect of sound
f = f₀ x (V - v₀) /(
)
f is apparent frequency ,f₀ is real frequency , V is velocity of sound , v₀ is velocity of observer going away ,
is velocity of source going away
778 = 840 x (340 - 14)/ (340 +
)
340 +
= 341.18
= 1.18 m /s
it will go away from the observer or the cyclist.
speed of train = 1.18 m /s
a )
For a stationary observer v₀ = 0
f = f₀ x V /(
)
= 840 x 340 / (340 + 1.180)
= 837 Hz
The solution for this problem:
Given:
f1 = 0.89 Hz
f2 = 0.63 Hz
Δm = m2 - m1 = 0.603 kg
The frequency of mass-spring oscillation is:
f = (1/2π)√(k/m)
k = m(2πf)²
Then we know that k is constant for both trials, we have:
k = k
m1(2πf1)² = m2(2πf2)²
m1 = m2(f2/f1)²
m1 = (m1+Δm)(f2/f1)²
m1 = Δm/((f1/f2)²-1)
m 1 = 0.603/
(0.89/0.63)^2 – 1
= 0.609 kg or 0.61kg or 610 g