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alexdok [17]
3 years ago
13

A 200-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. The total energy

of the system is 2.00 J. Find (a) the force constant of the spring and (b) the amplitude of the motion.
Physics
1 answer:
m_a_m_a [10]3 years ago
6 0

Answer:

(A) Spring constant will be 126.58 N/m

(B) Amplitude will be equal to 0.177 m

Explanation:

We have given mass of the block m = 200 gram = 0.2 kg

Time period T = 0.250 sec

Total energy is given TE = 2 J

(A) For mass spring system time period is equal to T=2\pi \sqrt{\frac{m}{K}}

So 0.250=2\times 3.14 \sqrt{\frac{0.2}{K}}

0.0398=\sqrt{\frac{0.2}{K}}

Now squaring both side

0.00158=\frac{0.2}{K}

K = 126.58 N/m

So the spring constant of the spring will be 126.58 N/m

(B) Total energy is equal to TE=\frac{1}{2}KA^2, here K is spring constant and A is amplitude

So 2=\frac{1}{2}\times 126.58\times A^2

A^2=0.0316

A = 0.177 m

So the amplitude of the wave will be equal to 0.177 m

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