The total energy of the body executing simple harmonic motion (SHM) is E. Then the kinetic energy when the displacement is half of the amplitude is

Question

The total energy of the body executing simple harmonic motion (SHM) is E. Then the kinetic energy when the displacement is half of the amplitude is (A)  (E/2)  (B)  (E/4)  (C)  (3E/4)  (D)  (√3E/4)

Tardigrade Admin · Accepted Answer

Total energy in SHM E=(1/2) m ω2 a2, (where a = amplitude) Kinetic energy K=(1/2) m ω2(a2-y2) =E-(1/2) m ω2 y2 when y=(a/2) ⇒ K=E-(1/2) m ω2((a2/4)) =E-(E/4) K=(3 E/4)

Q. The total energy of the body executing simple harmonic motion (SHM) is $E$ . Then the kinetic energy when the displacement is half of the amplitude is

$\frac{E}{2}$

$\frac{E}{4}$

$\frac{3 E}{4}$

$\frac{3 E}{4}$

Total energy in SHM $E = \frac{1}{2} m ω^{2} a^{2}$ ,
(where $a =$ amplitude)
Kinetic energy $K = \frac{1}{2} m ω^{2} (a^{2} - y^{2})$
$= E - \frac{1}{2} m ω^{2} y^{2}$
when $y = \frac{a}{2}$
$\Rightarrow K = E - \frac{1}{2} m ω^{2} (\frac{a ^{2}}{4})$
$= E - \frac{E}{4} K = \frac{3 E}{4}$

Q. The total energy of the body executing simple harmonic motion (SHM) is E. Then the kinetic energy when the displacement is half of the amplitude is

2E​

4E​

43E​

43E​​

Total energy in SHM E=21​mω2a2, (where a= amplitude) Kinetic energy K=21​mω2(a2−y2) =E−21​mω2y2 when y=2a​ ⇒K=E−21​mω2(4a2​) =E−4E​K=43E​

Q. The total energy of the body executing simple harmonic motion (SHM) is $E$ . Then the kinetic energy when the displacement is half of the amplitude is

$\frac{E}{2}$

$\frac{E}{4}$

$\frac{3 E}{4}$

$\frac{3 E}{4}$

Total energy in SHM $E = \frac{1}{2} m ω^{2} a^{2}$ ,
(where $a =$ amplitude)
Kinetic energy $K = \frac{1}{2} m ω^{2} (a^{2} - y^{2})$
$= E - \frac{1}{2} m ω^{2} y^{2}$
when $y = \frac{a}{2}$
$\Rightarrow K = E - \frac{1}{2} m ω^{2} (\frac{a ^{2}}{4})$
$= E - \frac{E}{4} K = \frac{3 E}{4}$