A spring having with a spring constant 1200 N m\\(\displaystyle ^{-1}\\) is mounted on a horizontal table as shown in the Figure. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

Spring constant, \\(k=1200\,m^{-1}\\)Mass, \\(m=3kg\\)Displacement, \\(A=2.0\,cm=0.02m\\)(i) Frequency of oscillation v, is given by the relation:\\(\displaystyle v=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\)where, T is time period\\(\therefore \displaystyle v=\frac{1}{2\times 3.14}\sqrt{\frac{1200}{3}} = 3.18 Hz\\)Hence, the frequency of oscillations is 3.18 cycles per second.(ii) Maximum acceleration (a) is given by the relation:\\(a=\omega^2\,A\\)where,\\(\omega =\\) Angular frequency \\(=\sqrt{\frac{k}{m}}\\)\\(A\\) = maximum displacement\\(\displaystyle \therefore a=\frac{k}{m}A=\frac{1200\times 0.02}{3}=8\, ms^{-2}\\)Hence, the maximum acceleration of the mass is \\(8.0\,m/s^2\\)(iii) Maximum velocity, \\(v_{max}=A\omega\\)\\(\displaystyle =A\sqrt{\frac{k}{m}}=0.02\times \sqrt{\frac{1200}{3}}=0.4\,m/s\\)Hence, the maximum velocity of the mass is 0.4 m/s.

Two masses \\(m_1\\) and \\(m_2\\) are suspended together by a massless spring of constant K as shown in figure. When the masses are in equilibrium, \\(m_1\\) is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of \\(m_2\\).

If only mass \\({ m }_{ 2 }\\) is present the extension of spring \\(l\\) is given by \\({ m }_{ 2 }g=kl\Rightarrow l=\cfrac { { m }_{ 2 }g }{ k } \\), where \\(k\\)=Spring constantNow with \\({ m }_{ 1 }\\) and \\({ m }_{ 2 }\\) the extension be \\(l'\\) \\(({ m }_{ 1 }+{ m }_{ 2 })g=kl'\\)\\(l'=\cfrac { { m }_{ 1 }+{ m }_{ 2 } }{ k } \\)When \\({ m }_{ 1 }\\) is removed, \\((l'-1)\\) acts as amplitude of oscillation\\(A=(l'-l)=(\cfrac { { m }_{ 1 }+{ m }_{ 2 } }{ k } )g-\cfrac { { m }_{ 2 }g }{ k } \\)\\(\Rightarrow A=\cfrac { { m }_{ 1 }g }{ k } \\)Now, if \\({ m }_{ 2 }\\) is removed the angular frequency of oscillation \\(\omega =\sqrt { \cfrac { k }{ { m }_{ 2 } } } \\)(This is same as the case of \\({ m }_{ 2 }\\) attach to spring)\\(*\\)N.B: When \\({ m }_{ 1 }\\) is removed the spring mass system will oscillates with the same angular frequency as the case of spring mass system with only mass \\({ m }_{ 2 }\\)

The frequency of vibration \\('f'\\) of a mass \\('m'\\) suspended from a spring of spring constant \\(K\\) is given by a relation of this type, \\(f = Cm^{x}K^{y}\\); where \\(C\\) is a dimensionless quantity. The value of \\(x\\) and \\(y\\) are

The frequency of oscillation spring mass system \\(=\cfrac { 1 }{ 2\pi } \sqrt { \cfrac { k }{ m } } \\)Where, \\(k\\)=spring constantand \\(m\\)=mass with springAnd given frequency, \\(f=c{ m }^{ x }{ k }^{ y }\\)Comparing above two frequency we get \\(x=-\cfrac { 1 }{ 2 } ,y=\cfrac { 1 }{ 2 } \\)

A spring balance has a scale that reads from \\(0\\) to \\(50 kg\\). The length of the scale is \\(20 cm\\). A body suspended from this balance, when displaced and released, oscillates with period of \\(0.6 s\\). What is the weight of the body ?

Maximum mass that the scale can read, \\(M=50kg\\)Maximum displacement of the spring = Length of the scale, \\(l=20cm = 0.2 m\\)Time period, \\(T=0.6s\\)Maximum force exerted on the spring, \\(F=Mg\\)where,\\(g=\\) acceleration due to gravity \\(=9.8\,m/s^2\\)\\(F=50\times 9.8=490\\)\\(\therefore\\) Spring constant, \\(k=F/l=490/0.2=2450\,Nm^{-1}\\)Mass m, is suspended from the balance.Times period, \\(t=2\pi\sqrt{\dfrac{m}{k}}\\)\\(\therefore m=\left(\dfrac{T}{2\pi}\right)^2\times k=\left(\dfrac{0.6}{2\times 3.14}\right)^2\times 2450=22.36kg\\)\\(\therefore\\) Weight of the body \\(=mg=22.36\times 9.8=219.16\,N\\)Hence, the weight of the body is about 219 N.

Two blocks \\(A\\) and \\(B\\) of masses \\(2m\\) and \\(m\\), respectively are connected by a massless and inextensible string. The whole system is syspended by a massless spring as shown in the figure. The magnitude of acceleration of \\(A\\) and \\(B\\) immediately after the string is cut, are respectively:

Two blocks A and B of mass \\(2m\\)and \\(m\\) respectively and they are connected by a mass less and in extensible string.The equilibrium condition FBD of \\(2m\\)\\(Kx=T+2mg\\)\\(=mg+2mg=3mg\\)After the string is cut\\(T=0\\)FBD of \\(2m\\)\\(a_2=\cfrac{3mg-2mg}{2m}\\ \quad=\cfrac{g}{2}\\)FBD of \\(m\\)Hence \\(\cfrac{g}{2}\\) and \\(g\\)

Two identical springs of constant are connected in series and parallel as shown in figure. A mass \\(m\\) is suspended from them. The ratio of their frequencies of vertical oscillations will be

We know that, \\(n=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\)In series, \\(k=\dfrac{k_1k_2}{k_1+k_2}=\dfrac{k^2}{2}\\)In parallel, \\(k=k_1+k_2=2k\\)\\(\therefore, n_1=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{2m}}\\)and \\(n_2 = \dfrac{1}{2\pi}\sqrt{\dfrac{2k}{m}}\\)hence, \\(n_1:n_2=1:2\\)

A spring of force constant \\(K\\) is cut into two pieces such that the one piece is double the length of the other. Then the long piece will have a force constant of:

Force constant, \\( K \propto \dfrac{1}{l}=\dfrac{c}{l}\\)Let the spring cut into pieces of length \\(x, 2x \\)So, \\(x+2x=l \Rightarrow x=\dfrac{l}{3}\\)Force constant of long piece is: \\( k'=\dfrac{c}{2x}=\dfrac{Kl}{2l/3}=\dfrac{3K}{2}\\)

A spring of spring constant 5 x \\(10 ^ { 3 }\\) N/m is etched initially by 5 cm from the unstretched position the work required to stretch it further by another 5 cm is

Work done is change in potential energyThe spring is initially stretch to 5 cm and again further stretched by 5 cm and further stretched by 5 cm\\(\begin{array}{l} w={ p_{ 10 } }-{ p_{ 5 } } \\ =\frac { 1 }{ 2 } \times 5\times { 10^{ 3 } }\left[ { I{ { \left( { \frac { { 10 } }{ { 100 } } } \right) }^{ 2 } }-{ { \left( { \dfrac { 5 }{ { 100 } } } \right) }^{ 2 } } } \right] \\ =\dfrac { 1 }{ 2 } \times 5\times { 10^{ 3 } }\left( { \dfrac { { 10 } }{ { 100 } } +\dfrac { 5 }{ { 100 } } } \right) \left( { \dfrac { { 10 } }{ { 100 } } -\dfrac { 5 }{ { 100 } } } \right) N-m \\ =\dfrac { 1 }{ 2 } \times 5\times { 10^{ 3 } }\times \dfrac { { 15 } }{ { 100 } } \times \dfrac { 5 }{ { 100 } } =2.5\times 15\times 5\times { 10^{ 3 } }\times { 10^{ -4 } } \\ =75\times 2.5\times { 10^{ -1 } }=7.5\times 2.5N-m \\ =18.75\, \, N-m \end{array}\\)Option A.

The time period of mass suspended from a spring is \\(T\\). If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be

\\(T = 2\pi \sqrt {\dfrac{m}{k}}\\)When the spring is cut into four equal parts, the spring constant of one part will becomes 4k, therefore the new time period will become\\(T' = 2\pi \sqrt {\dfrac{m}{k'}}\\)\\(T = 2\pi \sqrt {\dfrac{m}{4k}}\\) \\(T'=\dfrac{T}{2}\\)

A mass \\(M\\) is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period \\(T\\). If the mass is increased by \\(m\\), the time period becomes \\(\displaystyle\frac{5T}{3}\\). Then the ratio of \\(\displaystyle\frac{m}{M}\\) is

From formula 1, we can say\\(T=2\pi \sqrt{\dfrac{M}{k}}\\) ...(1)Also,\\(\dfrac{5T}{3}=2\pi \sqrt{\dfrac{M+m}{k}}\\) ...(2)Dividing eqn 2 and 1, we get\\(\dfrac{5}{3}= \sqrt{\dfrac{M+m}{M}}\\)\\(\therefore \dfrac{25}{9}= \dfrac{M+m}{M}\\)\\(\therefore \dfrac{25}{9}=1+ \dfrac{m}{M}\\)Hence, answer=16/9, i.e. option B

Oscillations Due to a Spring | Definition, Examples, Diagrams

example

Problem on velocity and amplitude of mass attched to spring

Example: A block with mass

M

is connected by a mass less spring with stiffness constant

k

to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position

x_{0}

. Consider two cases: (i) when the block is at

x_{0}

; and (ii) when the block is at

x = x_{0} + A

. In both the cases, a particle with mass

m (< M)

is softly placed on the block after which they stick to each other.
Which of the following statement(s) is(are) true about the motion after the mass

m

is placed on the mass

M

?

Solution:
In case I,From Conservation of momentum,

M V_{1} = (M + m) V_{2}

\frac{M V _{1}}{M + m} = V_{2}

\frac{k}{M + m} A_{2} = \frac{M}{M + m} \frac{k}{M} A_{1}

A_{2} = \frac{M}{M + m} A_{1}

In case II,

A_{2} = A_{1}

T = 2 π \frac{M + m}{k}

in both the cases.
Total energy decreases in first case where as remain same in 2nd case.
Instantaneous speed at

x_{0}

decreases in both case.

example

Write force equations of an arrangement of springs in parallel and calculate equivalent spring constant

Example: Two blocks

A (5 k g)

and

B (2 k g)

attached to the ends of a spring constant 1120N/m are placed on a smooth horizontal plane with the spring undeformed. Simultaneously, velocities of 3m/s and 10m/s along the line of the spring in the same direction are imparted to

A

and

B

, then find the maximum extension of the spring.

Solution:
The maximum elongation occurs when the velocity of both the particles becomes equal. Since there is no external force acting in the horizontal direction (for the
blocks + spring system), thus the linear momentum is conserved in horizontal direction:
If the final common velocity is

V

5 \times 3 + 2 \times 10 = (5 + 2) V V = 5 m s^{- 1}

Also applying law of conservation of energy for the system (work done by non-conservative force is zero):
Initial Energy

=

Final Energy

\frac{1}{2} \times 5 \times 3^{2} + \frac{1}{2} \times 2 \times 10^{2} = \frac{1}{2} \times 1120 x^{2} + \frac{1}{2} \times 7 \times 5^{2}

Solving the above equation we get

x = 25

cm
Also it can be observed that the compression and expansion occur alternately.

Oscillations Due to a Spring

example

Problem on velocity and amplitude of mass attched to spring

example

Write force equations of an arrangement of springs in parallel and calculate equivalent spring constant

REVISE WITH CONCEPTS

Problems on SHM of Springs

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