The figure shows a small block of mass m attached to a spring of force constant k and an identical spring hangs from the ceiling- Initially- lower spring is in the compressed state with compression equal to 3mg k from the natural length of spring- When the block is released it strikes upper spring and sticks to it- What is the amplitude of oscillation in cm after sticking given mg-10N-k-100-7N-m-

Mas =m Force constant =K Compression =3mgK mg=10N k=100√(7)N/m Amplitude of oscillation (in cm)After sticking given, #160; mg=kx ...

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Standard XII

Physics

Intro to Simple Pendulum

The figure sh...

Question

The figure shows a small block of mass $m$ attached to a spring of force constant $k$ and an identical spring hangs from the ceiling. Initially, lower spring is in the compressed state with compression equal to $\frac{3 m g}{k}$ from the natural length of spring. When the block is released it strikes upper spring and sticks to it. What is the amplitude of oscillation (in $c m$ ) after sticking given $m g = 10 N, k = 100 \sqrt{7} N / m$ .

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Solution

Mas $= m$
Force constant $= K$
Compression $= \frac{3 m g}{K}$
$m g = 10 N$
$k = 100 \sqrt{7} N / m$
Amplitude of oscillation (in cm)
After sticking given,
$m g = k x 10 = 100 \sqrt{7} \times x x = \frac{10}{100 \sqrt{7}} = \frac{1}{10 \sqrt{7}} c m$

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Mas=mForce constant=KCompression=3mgKmg=10Nk=100√7N/mAmplitude of oscillation (in cm)After sticking given, mg=kx10=100√7×xx=10100√7=110√7cm

Mas $= m$
Force constant $= K$
Compression $= \frac{3 m g}{K}$
$m g = 10 N$
$k = 100 \sqrt{7} N / m$
Amplitude of oscillation (in cm)
After sticking given,
$m g = k x 10 = 100 \sqrt{7} \times x x = \frac{10}{100 \sqrt{7}} = \frac{1}{10 \sqrt{7}} c m$