Question

A body is executing shm with amplitude a and time period t.the ratio of ke and pe when displacement from equilibrium position is half the amplitude a. 1:1b. 2:1c. 1:3d. 3:1​

Answer 1

Answer:

3:1

Explanation:

$K.E=\frac{1}{2} k(A^{2} - x^{2} ) \\\\P.E=\frac{1}{2} kx^{2}$

we have x =A/2

substituting in the above equations

$\frac{K.E}{P.E}= \frac{\frac{3}{8} kA^{2}}{\frac{1}{8} kA^{2}}$

there fore ans is 3:1

Answer 2

d.3:1

Explanation:

Initially amplitude=a

Time period=t

Kinetic energy,$k_e=\frac{1}{2}k(A^2-x^2)}$

Where k=Spring constant

A=Amplitude

x=Magnitude of displacement from equilibrium position

Potential energy,$P_e=\frac{1}{2}kx^2$

When x=$\frac{a}{2}$

Substitute the values

$K_e=\frac{1}{2}k(a^2-\frac{a^2}{4})=\frac{3}{8}ka^2$

$P_e=\frac{1}{2}k(\frac{a}{2})^2=\frac{1}{8}ka^2$

$\frac{K_e}{P_e}=\frac{\frac{3}{8}ka^2}{\frac{1}{8}ka^2}=\frac{3}{1}$

$K_e:P_e$=3:1

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