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frutty [35]
3 years ago
5

A block of unknown mass is attached to a spring of spring constant 9.4 N/m and undergoes simple harmonic motion with an amplitud

e of 15.5 cm. When the mass is halfway between its equilibrium position and the endpoint, its speed is measured to be 32.8 cm/s. Calculate the mass of the block. Answer in units of kg
Physics
1 answer:
g100num [7]3 years ago
5 0

Answer:

1.57 kg

Explanation:

Let the mass of block is m.

Spring constant, K = 9.4 N/m

Amplitude, A = 15.5 cm

y = A/2

v = 32.8 cm/s = 0.328 m/s

The velocity of the particle executing SHM is given by

v=\omega\sqrt{A^{2}-y^{2}}

where, ω is the angular frequency of SHM.

0.328=\omega\sqrt{A^{2}-\left ( \frac{A}{2} \right )^{2}}

0.328=\omega\times 0.866 A

0.328 = ω x 0.866 x 0.155

ω = 2.45 rad/s

Now the angular frequency is given by

\omega = \sqrt\frac{K}{m}

K = mω²

9.4 = m x 2.45 x 2.45

m = 1.57 kg  

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