A mass is set oscillating horizontally, with an amplitude of 11.3 cm and a period of 0.702 sec. What is the velocity of the particle when the position is 1/2 the amplitude in the positive x direction and moving toward x=0?

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Answer:

Explanation:

18.1

The appropriate solution is "-87.6 cm/s". A complete solution is given below.

According to the question, the given values are:

Amplitude,

a = 11.3 cm

Time,

t = 0.702 sec

→ The velocity (v) will be:

= [tex]\pm w\sqrt{a^2-x^2}[/tex]

or,

= [tex]-\frac{2 \pi}{t} \sqrt{a^2-(\frac{a}{2} )^2}[/tex]

= [tex]-\frac{2 \pi}{t} \frac{\sqrt{3} }{2}a[/tex]

By substituting the given values, we get

= [tex]-\frac{2 \pi}{0.702} \frac{\sqrt{3}\times 11.3 }{2}[/tex]

=  [tex]-8.946\times 9.7861[/tex]

= [tex]-87.6 \ cm/s[/tex]

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