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STALIN [3.7K]
2 years ago
8

A mass m0 is attached to a spring and hung vertically. The mass is raised a short distance in the vertical direction and release

d. The mass oscillates with a frequency f0 . If the mass is replaced with a mass nine times as large, and the experiment was repeated, what would be the frequency of the oscillations in terms of f0
Physics
1 answer:
iragen [17]2 years ago
6 0

Answer:

The frequency of the oscillations in terms of fo will be f2=fo/3

E xplanation:

T= 2pie\frac sqrt {m}{k}

 \frac {{f2}/times {fo}}=1:3

⇒f2=fo\3

Here frequency f is inversely poportional to square root of mass m.

so the value of remainder of frequency f2 and fo is equal to 1:3.

⇒\frac{f2} {f1} = \frac sqrt{m1}[m2}

⇒\frac{f2}{fo} = 1:3

⇒f2=\frac{fo} {3}

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Answer:

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Explanation:

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Initial velocity of the catcher, v_{ic} = 0~m/s

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&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s

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&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s

Thus, the final velocity of thrower is 3.88~m/s and that for the catcher is 0.029~m/s.

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