The frequency and amplitude of the SHM beam is 0.8 Hz and 0.098 m. The frequency of the SHM wave when gravel falls is 0.8 Hz and the amplitude of subsequent SHM beam is 0.4m.
(a) Mass of the spring = 225 Kg
Mass of the sack = 175 Kg
Amplitude of the beam = 40 cm = 0.40 m
Frequency of the beam = F = 0.60 cycles/s
The formula for frequency of oscillation =
= f = (1/2π) X √(k/m)
where, k = 2π²F²m
= k = 2 X (3.14)² X 0.6² X (225 + 175)
= k = 5685.37 N/m
Strength of the spring before gravels fall = x =
= x = mg / k
= x = [ (225 + 175 ) X 9.8 ] / 5685.37
= x = 0.689 m
But, the spring is stretched by the distance of x' which is expressed as,
= X = x - x'
= X = 0.689 - 0.40
= X = 0.289 m
Now, since we know that the gravel falls, thus frequency = f =
= f = (1/2π) X √(k/m)
= f= (1/ 2 X 3.14) X √ 5685.37 / 225
= f = 0.8 Hz
(b) Assuming that the spring is stretched, x = mg/k =
= x = (225 X 9.8) / 5685.37
= x = 0.3878 m
Thus, the amplitude of the sack = A = 0.3878 - 0.289
= A = 0.098 m
(c) If the gravel falls, the speed is maximum hence speed = s =
= s = A X √(k/m)
= s = 0.4 X √(5685.37/400)
= s = 1.508 m/s
The frequency = f' =
= f' = (1/2π) X √(k/m)
= f' = (1/2 X 2.14) X √(5685.37/225)
= f' = 0.8 Hz
(d) New amplitude = A' =
= A' = 0.38 + 0.038 (after calculating the new distance)
= A' = 0.4 m
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