As shown below, a 350g mass is hung vertically by a spring with an unknown sprin

Question

As shown below, a 350g mass is hung vertically by a spring with an unknown spring constant over a pool of still water. When the mass is brought to rest, the spring stretches 13.2 cm. The spring is stretched an additional 5.00 cm until the mass barely touches the surface of the water. Upon release, the oscillating mass creates sinusoidal waves in the water; the distance between adjacent crests is measured to be 32.0 cm. Assume there is no damping of the oscillator. Find the spring constant of the spring. Find the maximum speed of the mass. Find the speed of the water waves. Assume the spring of the system is replaced by another spring with twice the spring constant. Find the new wave speed, frequency, and wavelength of the water waves.

Explanation / Answer

part a:

let spring constant be k .

then at equilibrium, weight of the mass=k*elongation of the spring

==>0.35*9.8=k*0.132

==>k=25.9848 N/m

part b:

angular speed of oscilation=sqrt(k/m)=sqrt(25.9848/0.35)=8.6164 rad/s

maximum speed of the mass=angular speed*amplitude=8.6164*0.05=0.431 m/s

part c:

frequency of water waves=angular frequency/(2*pi)=1.37134 Hz

wavelength of water wave=distance between adjacent crests=32 cm=0.32 m

as we know, speed of a wave=frequency*wavelength=1.37134*0.32=0.43883 m/s

part d:new spring constant=2*k=51.9696 N/m

then new angular frequency=sqrt(51.9696/0.35)=12.1854 rad/s

then wave speed=angular frequency*amplitude=12.1854*0.05=0.60927 m/s

frequency=angular frequency/(2*pi)=1.93936 Hz

wavelength=speed/frequency=0.31416 m=31.416 cm

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