Question #58104

6 When a 30-g mass is hung from the end of a spring, the spring stretches 8.0 cm. The same spring with a mass of 200 g at at its end is stretched 5.0 cm, released and allowed to oscillate on a frictionless horizontal surface. Find the frequency of the oscillation.
0.54 Hz
0.68 Hz
0.34 Hz
9.5 Hz

7 The system shown is an example of the Atwood's machine. What is the acceleration of the masses? Assume the the pulley is frictionless and the rope massless. Take
g=9.8m/s2
4.2m/s2
7.4m/s2
9.8m/s2
3.3m/s2
1

Expert's answer

2016-03-18T15:48:03-0400

Answer on Question #58104 - Physics - Mechanics | Relativity

Task 6 When a 30-g mass is hung from the end of a spring, the spring stretches 8.0 cm. The same spring with a mass of 200 g at its end is stretched 5.0 cm, released and allowed to oscillate on a frictionless horizontal surface. Find the frequency of the oscillation.

0.54 Hz

0.68 Hz

0.34 Hz

9.5 Hz

Task 7 The system shown is an example of the Atwood's machine. What is the acceleration of the masses? Assume the pulley is frictionless and the rope massless. Take

g=9.8m/s2

4.2m/s2

7.4m/s2

9.8m/s2

3.3m/s2

Solution

Task 7. Can't be solved, because there is not enough data.

Task 6.

In first, due to Hooke's law system is in equilibrium, so that mg=kxmg = kx, where mm – is mass of 30g30g, gg – is free fall acceleration, kk – Hooke's coefficient for sprint, xx – spring extension. From here we can find kk.


k=mgx=0.03100.08=3.75Nmk = \frac{mg}{x} = \frac{0.03 \cdot 10}{0.08} = 3.75 \frac{N}{m}


Frequency of oscillation can be found as f=2πkMf = 2\pi \sqrt{\frac{k}{M}}, where MM is 200g=0.2kg200g = 0.2 \, \text{kg}

f=12π3.750.2=0.68Hzf = \frac{1}{2\pi} \sqrt{\frac{3.75}{0.2}} = 0.68 \, \text{Hz}


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