99 Percentile Qs Bank for JEE MainPhysicsOscillations
A body of mass ( I kg ) is suspended from a spring of negligible mass. Another body of mass (500 ~g ) moving vertically upwards hits the suspended body with a velocity of (3 ~ms ⁻¹ ) and gets embedded in it. If the frequency of oscillation of the system of the two bodies after collision is ( 10 Hz ), the amplitude of the motion and the spring constant are respectively,
Options
- A(5 ~cm , 300 Nm ⁻¹ )
- B(10 ~cm , 300 Nm ⁻¹ )
- C(10 ~cm , 600 Nm ⁻¹ )
- D(5 ~cm , 600 Nm ⁻¹ )
Correct answer
D. (5 ~cm , 600 Nm ⁻¹ )
Step-by-step solution
Given, mass of body, (M=1 ~kg ), frequency of oscillation, (f= 10 Hz ), mass of hitting body, (m=0.5 ~kg ) and speed of body, (v=3 ~ms ⁻¹ ) As frequency of oscillation of spring mass system, (f= 1 2 k M_ body ) Since, hitting body get embedded to the initial mass, so mass of the body changed to (M_ body =M+m=1+0.5=1.5 ~kg ) Spring constant, ( k=(2 /)^2 M_ body ) (= (2 10 )^2(1.5)=600 Nm ⁻¹ ) Now, velocity of body after collision, ( aligned & (M+m) v^ =m v (1+1 / 2) v^ =1 / 2 3 & v^ =1 ~m / s aligned ) The maximum a