Question

A uniform beam is suspended horizontally by two identical vertical springs that are attached between the ceiling and each end of the beam. the beam has mass 225 kg, and a 175-kg sack of gravel sits on the middle of it. the beam is oscillating in shm, with an amplitude of 40.0 cm and a frequency of 0.60 cycles/s.What is the frequency of the subsequent SHM of the beam?
1. What is the amplitude of the subsequent SHM of the beam?
2. If the gravel instead falls off when the beam has its maximum speed, what is the frequency of the subsequent SHM of the beam?
3. What is the amplitude of the subsequent SHM of the beam?

Answer 1

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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