⭕ Answer in detail, thanks!
Asked by jhajuhi19 | 12th Aug, 2021, 01:57: PM
Figure shows the oscillating mass that is attached with spring left side and attached with elastic string right side.
Let the mass be displaced towards right by a distance x from equilibrium position .
As shown in free body diagram , spring pulls the mass with force F1 = -K x.
Since the elstic tring becomes loose , there is no force acting on mass due to elastic string.
Once the mass is released, it returns back towards equilibrium position and crosses equilibrium position.
When it crosses equilibrium position, the mass compressing the spring and stretches the elastic string.
Now as shown in free body diagram, when the displacement is x , restoring force (-Kx ) from spring
pushes the mass and restoring force (-Kx) from elastic string pulls the mass.
Hence for the half cycle of oscillation from equilibrium position to right most point then back to equilibrium,
total restoring force acting on mass is (-Kx) .
Hence haif period TR of this part of cyle is given as,
For the next half cycle of oscillation from equilibrium position to left most point then back to equilibrium,
total restoring force acting on mass is (-2Kx) .
Hence haif period TL of this part of cyle is given as,
Full time period
Answered by Thiyagarajan K | 12th Aug, 2021, 07:57: PM
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