Question

The force on a body executing SHM is 4 N when
the displacement from mean position is 2 cm. If
amplitude of oscillation is 10 cm, then the
maximum kinetic energy associated with the SHM
will be
[NCERT Pg. 351]
(1) 1 J
(2) 4J
(3) 2 J
(4) 3J​

Answer 1

Answer:

1J

Explanation:

using the formula F= -kx

and max Kinetic energy = 1/2 kA^2

Answer 2

Answer:

The maximum energy associated with SHM will be 1 Joule i.e option (1).

Explanation:

Here we will require two expressions to solve the given problem and then we will get our answer

the first expression we need is,

F=kx     (1)

F=force

k=spring constant

x=displacement

the second expression is,  

K=1/2 × k ×$A^{2}$     (2)

K=kinetic energy

k=spring constant

A=amplitude

the given values in question are,

F=4N, x=2cm=2×$10^{-2}m$,A=10cm=10×$10^{-2} m$

for finding kinetic energy we first need to find k

so, we will  use equation 1

4=k x 2 x $10^{-2}$

k=200

now let's use equation 2 then we will get our final answer

K=1/2 x 200 x (10×$10^{-2}$$)^{2}$

K=1 Joule