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A point on the string of a violin moves up and down in simple harmonic motion with an amplitude of 1.24 mm and frequency of 875 Hz. a) what is the max speed of that point in SI units? b) what is the max acceleration of the point in SI units?

User Denise
by
4.8k points

2 Answers

3 votes

Answer:

a


v _(max ) = 6.82 \ m/s

b


a_(max) = 37489.5 \ m/s^2

Step-by-step explanation:

From the question we are told that

The amplitude is
A = 1.24 \ mm = 1.24 * 10^(-3) \ m

The frequency is
f = 875 \ Hz

Generally the maximum speed is mathematically represented as


v _(max ) = A * 2 * \pi * f

=>
v _(max ) = 1.24*10^(-3) * 2 * 3.142 * 875

=>
v _(max ) = 6.82 \ m/s

Generally the maximum acceleration is mathematically represented as


a_(max) = A * (2 * \pi * f)

=>
a_(max) = 1.24*10^(-3) * (2 * 3.142 * 875 )^2

=>
a_(max) = 37489.5 \ m/s^2

User Cory Trese
by
4.8k points
3 votes

Using

V = Amplitude x angular frequency(omega)

But omega= 2πf

= 2πx875

=5498.5rad/s

So v= 1.25mm x 5498.5

= 6.82m/s

B. .Acceleration is omega² x radius= 104ms²

User Josh Albert
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4.7k points