Question

he position of a mass-on-a-spring oscillator is given by

y = A sin(15t),

where the value of t is in seconds and

A = 0.34 m.

(a) What is the maximum kinetic energy of an oscillator of mass 2.2 kg?
J

(b) Suppose the amplitude is increased so that the maximum kinetic energy is doubled. What is the new value of A?
m


2) A guitar string of diameter 0.6 mm and length 0.69 m is subject to a tension of 146 N. If the string stretches an amount 0.41 mm, what is Young's modulus of the string?
Pa


3)The displacement of a harmonic oscillator is given by y = 2.2 sin(23t), where the units of y are meters and t is measured in seconds.

(a) Find the amplitude.
m

(b) Find the frequency.
Hz

Answer 1

1.

a) Kinetic energy is equal to 1/2*M*V^2. Here m (mass) is known which is 1.7Kg. now velocity of SHM or spring oscillator is w.sqrt(A^2-x^2) where w= angular frequency=15, A=amplitude=.34 so velocity will be max when displacement will be zero so max velocity = w.A=15*.34=5.1. So max
K.E. = 1/2 * M*V^2 = 1/2*2.2*5.1*5.1 = 28.611 J.

b) K.E. = 1/2 * M*V^2=K.E. = 1/2 * M*(A*w)^2
Now K.E. is proportional to square of Amplitude, hence when K.E. becom twice, Amplitude will be sqrt(2) times of A.

2.Y = stress/strain = F/A/(?L/L) = F*L/(A*?L) = 146N*0.69m/(?*(0.6x10^-3m)^2 x0. 41x10^-3m)

= 21.7x10^10Pa

3.y = 2.2 sin(23t)

You need to think of y = Asin(wt) type of equations where A is the amplitude and w is the angular frequency of the oscillation.

The max value of a sine function is +1 so the max value of y is going to be 2.2 x 1 = 2.2 and this is the Amplitude.

The min value of a sine function is -1 so the min value of y is going to be 2.2 x -1 = -2.2 and this is the smallest value for y.
You see that the wave oscillates between -2.2 and +2.2

Notice that w = 23 so that the time period is 2pi/w = 2pi/23 ~ 0.273 seconds.
the frequency is given by 1/T = 1/0.273 ~ 3.663 Hz

Hope this helps.