364373
A mass \(m\) is attached to two springs as shown in figure. The spring constants of two springs are \(K_{1}\) and \(K_{2}\). For the frictionless surface, the time period of oscillation of mass is
As we can see from given figure, both the springs are connected in parallel combination. \(\therefore K_{e q}=K_{1}+K_{2}\) As time period is \(T=2 \pi \sqrt{\dfrac{m}{K_{e q}}}\) \(\Rightarrow T=2 \pi \sqrt{\dfrac{m}{K_{1}+K_{2}}}\)
JEE - 2023
PHXI14:OSCILLATIONS
364374
As shown in figure a simple harmonic motion oscillator having identical four springs has time period
1 \(T=2 \pi \sqrt{\dfrac{m}{2 k}}\)
2 \(T=2 \pi \sqrt{\dfrac{m}{4 k}}\)
3 \(T=2 \pi \sqrt{\dfrac{m}{8 k}}\)
4 \(T=2 \pi \sqrt{\dfrac{m}{k}}\)
Explanation:
The upper two springs are in parallel their equival ent spring constant is \(2 K\). Similarly the equivalent spring constant for two lower springs is \(2\;K\). The equivalent spring constant of the system is \(\begin{aligned}& \dfrac{1}{k_{e q}}=\dfrac{1}{2 k}+\dfrac{1}{2 k} \Rightarrow k_{e q}=k \\& T=2 \pi \sqrt{\dfrac{m}{k}}\end{aligned}\)
PHXI14:OSCILLATIONS
364375
A mass \(m\) is suspended separately by two different springs in successive order, then time periods is \(t_{1}\) and \(t_{2}\) respectively. If \(m\) is connected by both springs as shown in figure, then time period is \(t_{0}\), the correct relation is
364376
In the figure all springs are identical having spring constant \(k\)and mass \(m\) each. The block also has mass \(m\). The frequency of oscillation of the block is:
1 \(\dfrac{1}{2 \pi} \sqrt{\dfrac{3 k}{2 m}}\)
2 \(\dfrac{1}{2 \pi} \sqrt{\dfrac{3 k}{m}}\)
3 \(2 \pi \sqrt{\dfrac{3 m}{3 k}}\)
4 None of these
Explanation:
If spring has the mass then: \(f=\dfrac{1}{2 \pi} \sqrt{\dfrac{k}{m+\dfrac{M_{\text {spring }}}{3}}}\) Spring equivalent \(=3 k\) (As all are in parallel \()\) \(f=\dfrac{1}{2 \pi} \sqrt{\dfrac{3 k}{m+\dfrac{(3 m)}{3}}} \Rightarrow f=\dfrac{1}{2 \pi} \sqrt{\dfrac{3 k}{2 m}}\)
364373
A mass \(m\) is attached to two springs as shown in figure. The spring constants of two springs are \(K_{1}\) and \(K_{2}\). For the frictionless surface, the time period of oscillation of mass is
As we can see from given figure, both the springs are connected in parallel combination. \(\therefore K_{e q}=K_{1}+K_{2}\) As time period is \(T=2 \pi \sqrt{\dfrac{m}{K_{e q}}}\) \(\Rightarrow T=2 \pi \sqrt{\dfrac{m}{K_{1}+K_{2}}}\)
JEE - 2023
PHXI14:OSCILLATIONS
364374
As shown in figure a simple harmonic motion oscillator having identical four springs has time period
1 \(T=2 \pi \sqrt{\dfrac{m}{2 k}}\)
2 \(T=2 \pi \sqrt{\dfrac{m}{4 k}}\)
3 \(T=2 \pi \sqrt{\dfrac{m}{8 k}}\)
4 \(T=2 \pi \sqrt{\dfrac{m}{k}}\)
Explanation:
The upper two springs are in parallel their equival ent spring constant is \(2 K\). Similarly the equivalent spring constant for two lower springs is \(2\;K\). The equivalent spring constant of the system is \(\begin{aligned}& \dfrac{1}{k_{e q}}=\dfrac{1}{2 k}+\dfrac{1}{2 k} \Rightarrow k_{e q}=k \\& T=2 \pi \sqrt{\dfrac{m}{k}}\end{aligned}\)
PHXI14:OSCILLATIONS
364375
A mass \(m\) is suspended separately by two different springs in successive order, then time periods is \(t_{1}\) and \(t_{2}\) respectively. If \(m\) is connected by both springs as shown in figure, then time period is \(t_{0}\), the correct relation is
364376
In the figure all springs are identical having spring constant \(k\)and mass \(m\) each. The block also has mass \(m\). The frequency of oscillation of the block is:
1 \(\dfrac{1}{2 \pi} \sqrt{\dfrac{3 k}{2 m}}\)
2 \(\dfrac{1}{2 \pi} \sqrt{\dfrac{3 k}{m}}\)
3 \(2 \pi \sqrt{\dfrac{3 m}{3 k}}\)
4 None of these
Explanation:
If spring has the mass then: \(f=\dfrac{1}{2 \pi} \sqrt{\dfrac{k}{m+\dfrac{M_{\text {spring }}}{3}}}\) Spring equivalent \(=3 k\) (As all are in parallel \()\) \(f=\dfrac{1}{2 \pi} \sqrt{\dfrac{3 k}{m+\dfrac{(3 m)}{3}}} \Rightarrow f=\dfrac{1}{2 \pi} \sqrt{\dfrac{3 k}{2 m}}\)
364373
A mass \(m\) is attached to two springs as shown in figure. The spring constants of two springs are \(K_{1}\) and \(K_{2}\). For the frictionless surface, the time period of oscillation of mass is
As we can see from given figure, both the springs are connected in parallel combination. \(\therefore K_{e q}=K_{1}+K_{2}\) As time period is \(T=2 \pi \sqrt{\dfrac{m}{K_{e q}}}\) \(\Rightarrow T=2 \pi \sqrt{\dfrac{m}{K_{1}+K_{2}}}\)
JEE - 2023
PHXI14:OSCILLATIONS
364374
As shown in figure a simple harmonic motion oscillator having identical four springs has time period
1 \(T=2 \pi \sqrt{\dfrac{m}{2 k}}\)
2 \(T=2 \pi \sqrt{\dfrac{m}{4 k}}\)
3 \(T=2 \pi \sqrt{\dfrac{m}{8 k}}\)
4 \(T=2 \pi \sqrt{\dfrac{m}{k}}\)
Explanation:
The upper two springs are in parallel their equival ent spring constant is \(2 K\). Similarly the equivalent spring constant for two lower springs is \(2\;K\). The equivalent spring constant of the system is \(\begin{aligned}& \dfrac{1}{k_{e q}}=\dfrac{1}{2 k}+\dfrac{1}{2 k} \Rightarrow k_{e q}=k \\& T=2 \pi \sqrt{\dfrac{m}{k}}\end{aligned}\)
PHXI14:OSCILLATIONS
364375
A mass \(m\) is suspended separately by two different springs in successive order, then time periods is \(t_{1}\) and \(t_{2}\) respectively. If \(m\) is connected by both springs as shown in figure, then time period is \(t_{0}\), the correct relation is
364376
In the figure all springs are identical having spring constant \(k\)and mass \(m\) each. The block also has mass \(m\). The frequency of oscillation of the block is:
1 \(\dfrac{1}{2 \pi} \sqrt{\dfrac{3 k}{2 m}}\)
2 \(\dfrac{1}{2 \pi} \sqrt{\dfrac{3 k}{m}}\)
3 \(2 \pi \sqrt{\dfrac{3 m}{3 k}}\)
4 None of these
Explanation:
If spring has the mass then: \(f=\dfrac{1}{2 \pi} \sqrt{\dfrac{k}{m+\dfrac{M_{\text {spring }}}{3}}}\) Spring equivalent \(=3 k\) (As all are in parallel \()\) \(f=\dfrac{1}{2 \pi} \sqrt{\dfrac{3 k}{m+\dfrac{(3 m)}{3}}} \Rightarrow f=\dfrac{1}{2 \pi} \sqrt{\dfrac{3 k}{2 m}}\)
364373
A mass \(m\) is attached to two springs as shown in figure. The spring constants of two springs are \(K_{1}\) and \(K_{2}\). For the frictionless surface, the time period of oscillation of mass is
As we can see from given figure, both the springs are connected in parallel combination. \(\therefore K_{e q}=K_{1}+K_{2}\) As time period is \(T=2 \pi \sqrt{\dfrac{m}{K_{e q}}}\) \(\Rightarrow T=2 \pi \sqrt{\dfrac{m}{K_{1}+K_{2}}}\)
JEE - 2023
PHXI14:OSCILLATIONS
364374
As shown in figure a simple harmonic motion oscillator having identical four springs has time period
1 \(T=2 \pi \sqrt{\dfrac{m}{2 k}}\)
2 \(T=2 \pi \sqrt{\dfrac{m}{4 k}}\)
3 \(T=2 \pi \sqrt{\dfrac{m}{8 k}}\)
4 \(T=2 \pi \sqrt{\dfrac{m}{k}}\)
Explanation:
The upper two springs are in parallel their equival ent spring constant is \(2 K\). Similarly the equivalent spring constant for two lower springs is \(2\;K\). The equivalent spring constant of the system is \(\begin{aligned}& \dfrac{1}{k_{e q}}=\dfrac{1}{2 k}+\dfrac{1}{2 k} \Rightarrow k_{e q}=k \\& T=2 \pi \sqrt{\dfrac{m}{k}}\end{aligned}\)
PHXI14:OSCILLATIONS
364375
A mass \(m\) is suspended separately by two different springs in successive order, then time periods is \(t_{1}\) and \(t_{2}\) respectively. If \(m\) is connected by both springs as shown in figure, then time period is \(t_{0}\), the correct relation is
364376
In the figure all springs are identical having spring constant \(k\)and mass \(m\) each. The block also has mass \(m\). The frequency of oscillation of the block is:
1 \(\dfrac{1}{2 \pi} \sqrt{\dfrac{3 k}{2 m}}\)
2 \(\dfrac{1}{2 \pi} \sqrt{\dfrac{3 k}{m}}\)
3 \(2 \pi \sqrt{\dfrac{3 m}{3 k}}\)
4 None of these
Explanation:
If spring has the mass then: \(f=\dfrac{1}{2 \pi} \sqrt{\dfrac{k}{m+\dfrac{M_{\text {spring }}}{3}}}\) Spring equivalent \(=3 k\) (As all are in parallel \()\) \(f=\dfrac{1}{2 \pi} \sqrt{\dfrac{3 k}{m+\dfrac{(3 m)}{3}}} \Rightarrow f=\dfrac{1}{2 \pi} \sqrt{\dfrac{3 k}{2 m}}\)
