First slide

Applications of SHM

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.

Moderate

Solution

$T = 2 π \sqrt{\frac{m}{K_{1} + K_{2}}} = 2 π \sqrt{\frac{10}{360}} = \frac{π}{3} s$
The maximum velocity is always at equilibrium position since at any other point there will be a restoring force attempting to slow the mass.
$\begin{array}{l} ∴ V_{max} = \frac{impulse}{mass} = \frac{50}{10} = 5 {ms}^{- 1} \\ A = amplitude = \frac{V_{max}}{ω} = \frac{5}{\frac{(2 π)}{(\frac{π}{3})}} = \frac{5}{6} \Rightarrow δ = 6 \end{array}$

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