Question 1:
moderateA body is executing simple harmonic motion. At a displacement x, its potential energy is E1 and at a displacement y, its potential energy is E2. The potential energy E at a displacement (x + y) is :
\[ E_{1}= \frac{1}{2}Kx^{2} \]
\[ E_{2}= \frac{1}{2}Ky^{2} \]
\[Â E= \frac{1}{2}K(x+y)^{2}= \frac{1}{2}Kx^{2} + \frac{1}{2}Ky^{2} + Kxy \]
Question 2:
moderateA block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 Nm–¹. The block is pulled to a distance of x = 10 cm from its equilibrium position at x = 0 cm on a frictionless surface from rest at t = 0. The kinetic energy of the block when it is 5 cm away from the mean position isÂ
\[ K.E= \frac{1}{2}K\left( A^{2}-x^{2} \right)\]
\[ K.E= \frac{1}{2}\times 50\left( 10^{2}-5^{2} \right)/10^{4}= 0.19 J \]
Question 3:
moderateThe total mechanical energy of a spring-mass system in simple harmonic motion is E=1/2mω²A². Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude A remains the same. The new mechanical energy will :
Question 4:
moderateThe total mechanical energy of a particle executing simple harmonic motion is E. When the displacement is half the amplitude its kinetic energy will be :
\[ K.E= \frac{1}{2}K\left(A^{2}-\left( \frac{A}{2}\right)^{2} \right)= \frac{3}{8} K A^{2} \]
K.E= 3E/4
Question 5:
moderateA particle is executing S.H.M., If its P.E. & K.E. is equal then the ratio of displacement & amplitude will be :
\[ K.E= \frac{1}{2}K\left(A^{2}-x^{2} \right) \]
\[ P.E= \frac{1}{2}Kx^{2} \]
\[ \frac{1}{2}K\left(A^{2}-x^{2} \right)= \frac{1}{2}Kx^{2} \]
\[ x= A/\sqrt{2} \]
Question 6:
moderateA particle is executing linear simple harmonic motion of amplitude A. What fraction of the total energy is kinetic when the displacement is half the amplitudeÂ
\[ K.E= \frac{1}{2}K\left(A^{2}-x^{2} \right) \]
\[ K.E= \frac{1}{2}K\left(A^{2}-\left( \frac{A}{2}\right)^{2} \right)= \frac{3}{8} K A^{2} \]
K.E/T.E =3/4
Question 7:
moderateA particle of mass 0.1 kg executes SHM under a force F = (–10x) Newton. Speed of particle at mean position is 6 m/s. Then amplitude of oscillations is :
Spring Constant K = m ω² ⇒ 10 = 0.1 ω² ⇒ ω²= 100 ⇒ ω = 10 rad/sec
Speed is maximum at mean positionÂ
Vmax= Aω
6= A × 10
A = 0.6 m
Question 8:
moderateThe potential energy of a particle executing simple harmonic motion at a distance x from the equilibrium position is proportional to :
\[ U = \frac{1}{2}kx^{2} \]
Question 9:
moderateFor the damped oscillator, if time taken for its amplitude of vibrations to drop to half of its initial value is \(T\) then time taken for amplitude to drop to one eighth amplitude is
Amplitude decays exponentially as \(A = A_0 e^{-\gamma t}\). Since it halves in time \(T\), to drop to \(1/8 = (1/2)^3\) of its initial value, it takes exactly \(3T\).
Question 10:
moderateThe particle executing simple harmonic motion has a kinetic energy $K_0 \cos^2 \omega t$. The maximum values of the potential energy and the total energy are respectively:
(2007)
The maximum kinetic energy is $K_0$.\nIn an ideal SHM without damping, the total energy remains conserved and equals the maximum kinetic energy, which is $K_0$.\nThe maximum potential energy is also equal to the total energy, which is $K_0$.