A body of mass $1 k g$ is suspended from a weightless spring having force constant $600 N / m$ . Another body of mass $0.5 k g$ moving vertically upwards hits the suspended body with a velocity of $3.0 m / s$ and get embedded in it. Find the frequency of oscillations and amplitude of motion.

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Solution

Velocity of combined is $v_{1}$
By energy conservation,
$(M + m) v_{1} = m v$
$1.5 v_{1} = 0.5 \times 3$
$v_{1} = 1 m / s$
By energy conservation,
$\frac{1}{2} k A^{2} = \frac{1}{2} (M + m) v_{1}^{2}$
Frequency of oscillation is,
$ω = \sqrt{\frac{k}{M + m}}$
Amplitude $A = \sqrt{\frac{M + m}{k} .} v_{1}$
$= \frac{1}{20} \times 1 = 0.05 m = 5 c m$

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A body of mass 1kg is suspended from a weightless spring having force constant 600N/m. Another body of mass 0.5kg moving vertically upwards hits the suspended body with a velocity of 3.0m/s and get embedded in it. Find the frequency of oscillations and amplitude of motion.

Velocity of combined is v1By energy conservation,(M+m)v1=mv1.5v1=0.5×3v1=1m/sBy energy conservation,12kA2=12(M+m)v21Frequency of oscillation is,ω=√kM+mAmplitude A=√M+mk.v1=120×1=0.05m=5cm

A body of mass $1 k g$ is suspended from a weightless spring having force constant $600 N / m$ . Another body of mass $0.5 k g$ moving vertically upwards hits the suspended body with a velocity of $3.0 m / s$ and get embedded in it. Find the frequency of oscillations and amplitude of motion.