Two springs with negligible masses and force constant of $k_{1} = 200 {Nm}^{- 1} and k_{2} = 160 {Nm}^{- 1}$ are attached to the block of mass m = 10 kg as shown in the figure. Initially the block is at rest, at the equilibrium position in which both springs are neither stretched nor compressed. At time t =0, sharp impulse of 50 Ns is given to the block with a hammer along the spring.
Calculate $δ$ if the amplitude of SHM is (5/ $δ$ ) m.

Question

Two springs with negligible masses and force constant of $k_{1} = 200 {Nm}^{- 1} and k_{2} = 160 {Nm}^{- 1}$ are attached to the block of mass m = 10 kg as shown in the figure. Initially the block is at rest, at the equilibrium position in which both springs are neither stretched nor compressed. At time t =0, sharp impulse of 50 Ns is given to the block with a hammer along the spring.

Calculate $δ$ if the amplitude of SHM is (5/ $δ$ ) m.

Infinity Learn · Accepted Answer

$$T=2$$$$π$$$$mK1+K2=2$$$$π$$$$10360=$$π$$3sThe$$ maximum velocity is always at equilibrium position since at any other point there will be a restoring force attempting to slow the mass.∴ $$Vmax=$$ impulse  mass $$=5010=5ms$$−$$1A=$$ amplitude $$=Vmax$$ω$$=5(2$$π$$)$$π$$3=56$$⇒δ$$=6$$

Applications of SHM

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Applications of SHM

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T=2πmK1+K2=2π10360=π3sThe maximum velocity is always at equilibrium position since at any other point there will be a restoring force attempting to slow the mass.∴ Vmax= impulse mass =5010=5ms−1A= amplitude =Vmaxω=5(2π)π3=56⇒δ=6

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