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Shm lo hayley sinclair

H
Hayley Sinclair

The document describes the velocity of an A string on a violin after 2 seconds of oscillation. It is given that the A string has a frequency of 440 Hz, an initial position of /2, and an amplitude of 1.0 mm. The document solves for the unknown period and angular frequency, then uses the simple harmonic motion equation to find that the velocity of the string after 2 seconds is 2.76460 m/s or 2760 mm/s.

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Simple Harmonic Motion of a Violin String Open A on the Violin has a frequency of 440 Hz. If the A string is plucked so its initial location of oscillation is /2, and it oscillates with an amplitude of 1.0mm, what would the velocity of the string be after 2.0 seconds? ! We know that the equation for Velocity is: V(t)=(A)sin(t+) 1) Write out your knowns A= 1mm, or 0.001 m t= 2.0 s =/2 f= 440 Hz T=? =? ! !
2) Find your unknowns ! The period, T, can be found with the equation T=1/f, and because we know that f is 440Hz, its easy to solve for the period! T=1/440 ! ! is equal to 2/T, so just sub in our value for T. = 2/(1/440) ! 3) Sub into our equation for velocity ! V(t)=(A)sin(t+) V(2.0s)= [(2/(1/440)rad/second)(0.001 m)] x sin[(2/(1/440)rad/second)(2.0s)+(/2)] ! This gives us V(2.0)=2.76460 m/s OR 2760 mm/s..Thats pretty fast! ! 4) Rationalize your answer Look at a simple graph of Sin, try to think of all of the variables we are using in terms of this graph. describes different possible initial locations of the oscillation (in our example this is /2) and A is the amplitude of the curve (1 mm). We Pulled the string 1 mm, to its position of /2 to start the oscillation. !

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Shm lo hayley sinclair

  • 1. Simple Harmonic Motion of a Violin String Open A on the Violin has a frequency of 440 Hz. If the A string is plucked so its initial location of oscillation is /2, and it oscillates with an amplitude of 1.0mm, what would the velocity of the string be after 2.0 seconds? ! We know that the equation for Velocity is: V(t)=(A)sin(t+) 1) Write out your knowns A= 1mm, or 0.001 m t= 2.0 s =/2 f= 440 Hz T=? =? ! !
  • 2. 2) Find your unknowns ! The period, T, can be found with the equation T=1/f, and because we know that f is 440Hz, its easy to solve for the period! T=1/440 ! ! is equal to 2/T, so just sub in our value for T. = 2/(1/440) ! 3) Sub into our equation for velocity ! V(t)=(A)sin(t+) V(2.0s)= [(2/(1/440)rad/second)(0.001 m)] x sin[(2/(1/440)rad/second)(2.0s)+(/2)] ! This gives us V(2.0)=2.76460 m/s OR 2760 mm/s..Thats pretty fast! ! 4) Rationalize your answer Look at a simple graph of Sin, try to think of all of the variables we are using in terms of this graph. describes different possible initial locations of the oscillation (in our example this is /2) and A is the amplitude of the curve (1 mm). We Pulled the string 1 mm, to its position of /2 to start the oscillation. !