A particle undergoes simple harmonic motion having time period T. The time taken in 3/8th oscillation is

³⁄₈ T
⁵⁄₈ T
⁵⁄₁₂ T
⁷⁄₁₂ T

Solution

A boy 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 displacement (x + y) is

√E = √E₁ - √E₂
√E = √E₁ + √E₂
E = E₁ + E₂
E = E₁ - E₂

Solution

A particle of mass 10 gm is describing S.H.M. along a straight line with period of 2 sec and amplitude of 10 cm. Its kinetic energy when it is at 5 cm from its equilibrium position is

37.5 π² erg
3.75 π² erg
375 π² erg
0.375 π² erg

Solution

The time period of the oscillating system (see figure) is

\(T=2\pi \sqrt{\frac{m}{k_{1}k_{2}}}\)
\(T=2\pi \sqrt{\frac{m}{k_{1}+k_{2}}}\)
\(T=2\pi \sqrt{mk_{1}+k_{2}}\)
None of these

Two wires are kept tight between the same pair of supports.The tensions in the wires are in the ratio 2 : 1, the radii are in the ratio 3 :1 and the densities are in the ratio 1 : 2. The ratio of their fundamental frequencies is

2 : 3
2 : 4
2 : 5
2 : 6

Solution

Two oscillators are started simultaneously in same phase.After 50 oscillations of one, they get out of phase by π,that is half oscillation. The percentage difference of frequencies of the two oscillators is nearest to

2%
1%
0.5%
0.25%

Solution

Phase change π in 50 oscillations.

Phase change 2π in 100 oscillations.

So frequency different ~ 1 in 100.

If the mass shown in figure is slightly displaced and then let go, then the system shall oscillate with a time period of

\(2\pi \sqrt{\frac{m}{3k}}\)
\(2\pi \sqrt{\frac{3m}{2k}}\)
\(2\pi \sqrt{\frac{2m}{3k}}\)
\(2\pi \sqrt{\frac{3k}{m}}\)

Solution

Three masses of 500 g, 300 g and 100 g are suspended at the end of a spring as shown, and are in equilibrium. When the 500 g mass is removed, the system oscillates with a period of 2 second. When the 300 g mass is also removed,it will oscillate with a period of

Solution

If a simple pendulum of length l has maximum angular displacement θ, then the maximum K.E. of bob of mass m is

¹⁄₂ml/g
mg/2l
mgl(1 - cosθ)
mgl sinθ/2

Solution

When the bob moves from maximum angular displacement θ to mean position, then the loss of gravitational potential energy is mgh

where h = l(1 - cosθ)

The total mechanical energy of a spring-mass system in simple harmonic motion is E = ¹⁄₂ mω²A². 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

become 2E
become E/2
become √2E
remain E

Solution

E = ¹⁄₂ m ^k⁄_m A² ⇒ E = ¹⁄₂ KA²

⇒ E does not depend on m

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