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Simple Harmonic Motion One end of a light elastic string of natural length l, passing...
Simple Harmonic Motion
One end of a light elastic string of natural length l, passing through a small smooth ring of mass m, is attached to a point O on the ceiling of the room. A particle P of mass M attached to the other end of the string hangs in equilibrium, with the ring being held at rest at the point O. If 2Mg is the modulus of elasticity of the string, the extension of the string in the equilibrium position is l /2.
The ring, now released from rest at O, slides vertically downward along the string, under gravity, collides and coalesces with P. The composite body consisting of the ring and the particle moves vertically downward with velocity `m / (M + m)* sqrt (3gl) ` .
Q: Show that the composite body performs simple harmonic motion with frequency `sqrt ((2Mg) / ((m + M)l))` .
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The composite body made up of the ring and the particle P performs simple harmonic motion (motion of a spring out of equilibrium) due to the smooth change of velocity downwards, as the mass at the end falls due to gravity, and then upwards as the tension in the string retracts it. At the maximum and minimum points of the oscillation, the force acting in the other direction to motion overcomes the force acting in the same direction.
The frequency of simple harmonic motion is given by
`omega = sqrt(k/mu)` where `k` is the 'spring constant' and `mu` the mass at the end of the spring.
The spring constant in turn is calculated using the formula
`k = F/x_(max)` where `x_(max)` is the maximum displacement of the spring/string and `F` is the force acting downwards on the spring/string.
Now, `x_(max)` is the displacement when the conjoined body of the ring and the particle P and the string are in equilibrium - the tension T equals the downward force. Using Hooke's Law
`T = (lambda/l)x_(max)`
At that point ` `T is equal to the downward force due to gravity (m+M)g and we have that the modulus of elasticity `lambda = 2Mg` (constant). Therefore,
`x_(max) = (Tl)/lambda = ((m+M)gl)/(2Mg) = ((m+M)/M)(l/2)`
Recalling that the spring constant is given by
`k = F/x_(max)` where ` ` F is the force acting downwards on the spring (due to gravity), then we have that
`k = ((m+M)g)(M/(m+M))(2/l) = (2Mg)/l`
Recalling further that the frequency of the composite body made up of the ring and the particle P is given by
`omega = sqrt(k/mu)` we have finally then that
`w = sqrt((2Mg)/((m+M)l))` answer as required
Posted by mathsworkmusic on August 23, 2013 at 12:36 PM (Answer #1)
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