Question

• A 10-gram mass is attached to a
horizontal spring with spring constant
• k = 1.21 N/m. It is pulled 5 cm out from
its equilibrium position
• and released at t = 0.Write x(t).​

Answer 1

Answer:

Angular frequency of S.H.M= √k\m= 5√2

Here, the body is displaced to maximum at x=10cm=0.1m. So, it is the amplitude of the motion velocity of the body performing S.H.M at any position x then mean position is v=ω √a^2-x^2

Here a=0.1m

x=5cm=0.05m

⇒v=5

2

(

0.01−0.0005

)

v=0.61m/s

Kinetic energy at this point=

2

1

mv

2

=

2

1

×1×(0.61)

2

≃0.19J