A load of mass m falls from a height h on to the scale pan hung from the spring as shown in the figure. If the spring constant is k and mass of the scale pan is zero and the mass m does not bounce relative to the pan

Correct answer: B. mg k ( 1+2 hk mg ) According to energy conservation principle, If, x_ 1 is maximum elongation in the spring when the particle is in its lowest extreme position. Then, array l mgh=(1)/(2) kx_ 1 ^ 2 -mgx_ 1 (1)/(2) kx_ 1 ^ 2 -mgx_ 1 -mgh=0 or, x_ 1 ^ 2 - 2 mg k x_ 1 - 2 mg k · h=0 x_ 1 = 2 mg k ± .( 2 mg k )^ 2 +4 × 2 mg k h] 2 array Amplitude A=X_ 1 -X_ 0 (elongation in spring for equilibrium position) A= mg k (1+ 2 hk mg )

BITSAT 2015

Exam: BITSAT
Year: 2015
Subject: Physics
Chapter: Laws of Motion

BITSAT 2015 · Physics · Laws of Motion

A load of mass m falls from a height h on to the scale pan hung...

A load of mass m falls from a height h on to the scale pan hung from the spring as shown in the figure. If the spring constant is

\mathrm{k}

and mass of the scale pan is zero and the mass

\mathrm{m}

does not bounce relative to the pan, then the amplitude of vibration is A load of mass m falls from a height h on to the scale pan hung from the spring as shown in the figure. If the spring

A load of mass m falls from a height h on to the scale pan hung from the spring as shown in the figure. If the spring

Options

A $\mathrm{mg} / \mathrm{d}$
B\frac{\mathrm{mg}}{\mathrm{k}} \sqrt{ $\left(\frac{1+2 \mathrm{hk}}{\mathrm{mg}}\right)}$
C\frac{\mathrm{mg}}{\mathrm{k}}+\frac{\mathrm{mg}}{\mathrm{k}} \sqrt{ $\left(\frac{1+2 \mathrm{hk}}{\mathrm{mg}}\right)}$
D\frac{\mathrm{mg}}{\mathrm{k}} \sqrt{ $\left(\frac{1+2 \mathrm{hk}}{\mathrm{mg}}-\frac{\mathrm{mg}}{\mathrm{k}}\right)}$

Correct answer

B. \frac{\mathrm{mg}}{\mathrm{k}} \sqrt{ $\left(\frac{1+2 \mathrm{hk}}{\mathrm{mg}}\right)}$

Step-by-step solution

According to energy conservation principle, If,

\mathrm{x}_{1}

is maximum elongation in the spring when the particle is in its lowest extreme position. Then,

\begin{array}{l}\mathrm{mgh}=\frac{1}{2} \mathrm{kx}_{1}^{2}-\mathrm{mgx}_{1}\\ \Rightarrow \frac{1}{2} \mathrm{kx}_{1}^{2}-\mathrm{mgx}_{1}-\mathrm{mgh}=0 \\ \text{ or, } \mathrm{x}_{1}^{2}-\frac{2 \mathrm{mg}}{\mathrm{k}} \mathrm{x}_{1}-\frac{2 \mathrm{mg}}{\mathrm{k}} \cdot \mathrm{h}=0 \\ \therefore \mathrm{x}_{1}=\frac{\frac{2 \mathrm{mg}}{\mathrm{k}} \pm \sqrt{\left.\left(\frac{2 \mathrm{mg}}{\mathrm{k}}\right)^{2}+4 \times \frac{2 \mathrm{mg}}{\mathrm{k}} \mathrm{h}\right]}}{2} \end{array}

Amplitude

\mathrm{A}=\mathrm{X}_{1}-\mathrm{X}_{0}

(elongation in spring for equilibrium position) \mathrm{A}=\frac{\mathrm{mg}}{\mathrm{k}} \sqrt{\left(1+\frac{2 \mathrm{hk}}{\mathrm{mg}}\right)}

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